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There are already enough answers but here is a rewrite of Poonen's and De Bruyn's answer in the language of wreath products.

Let $A$ be a group acting freely on the right of a finite set $X$. Then $\mathrm{Aut}_A(X)$ is well known to be isomorphic to the permutational wreath product $A\wr \mathrm{Sym}(X/A)=A^{X/A}\rtimes \mathrm{Sym}(X/A)$. The isomorphism depends on choosing a transversal for $X/A$ but any two isomorphisms are conjugate by an element of $A^{X/A}$. In concrete terms, choose a section $t\colon X/A\to X$ and send $f\in \mathrm{Aut}_A(X)$ to $(F,f')$ where $f'$ is the permutation of $X/A$ induced by $f$ and $F\colon X/A\to A$ is defined on $e\in X/A$ by $f(t(e))=t(f'(e))F(e)$. If $t'\colon X/A\to X$ is another section, then there is $h\in A^{X/A}$ such that $t'(e)=t(e)h(e)$ for all $e\in X/A$ and the isomorphism constructed from $t'$ is conjugate to the one constructed from $t$ by $(h,1)\in A\wr \mathrm{Sym}(X/A)$. I didn't give the inverse of the isomorphism because it is not needed to construct the sign map.

In the case $A$ is abelian, there is a homomorphism $\tau_0\colon A^{X/A}\rtimes \mathrm{Sym}(X/A)\to A$ given by $(f,\sigma)\mapsto \prod_{e\in X/A}f(e)$ (sometimes called the transfer map). This gives a homomorphism $\tau\colon \mathrm{Aut}_A(X)\to A$ which is independent of the isomorphism constructed above since isomorphisms coming from different sections (i.e. transversals) differ by an inner automorphism and $A$ is abelian.

Now consider the group $A=C_2$ (the two element cyclic group) and consider the set $E$ of distinct pairs of elements from $[n]=\{1,\ldots, n\}$. Then $A$ acts freely on the right of $E$ by having the nontrivial element switching the coordinates and the action of $S_n$ on $E$ commutes with this. We then the composed homomorphism $S_n\to \mathrm{Aut}_{C_2}(E)\xrightarrow{\,\tau\,} C_2$.

As usual we just then need to check the transposition $(1,2)$ is sent to the nontrivial element of $C_2$, but this follows easily if you choose as your transversal the ordered pairs $(i,j)$ with $i<j$ and follow through the construction.

There are already enough answers but here is a rewrite of Poonen's and De Bruyn's answer in the language of wreath products.

Let $A$ be a group acting freely on the right of a finite set $X$. Then $\mathrm{Aut}_A(X)$ is well known to be isomorphic to the permutational wreath product $A\wr \mathrm{Sym}(X/A)=A^{X/A}\rtimes \mathrm{Sym}(X/A)$. The isomorphism depends on choosing a transversal for $X/A$ but any two isomorphisms are conjugate by an element of $A^{X/A}$. In concrete terms, choose a section $t\colon X/A\to X$ and send $f\in \mathrm{Aut}_A(X)$ to $(F,f')$ where $f'$ is the permutation of $X/A$ induced by $f$ and $F\colon X/A\to A$ is defined on $e\in X/A$ by $f(t(e))=t(f'(e))F(e)$. If $t'\colon X/A\to X$ is another section, then there is $h\in A^{X/A}$ such that $t'(e)=t(e)h(e)$ for all $e\in X/A$ and the isomorphism constructed from $t'$ is conjugate to the one constructed from $t$ by $(h,1)\in A\wr \mathrm{Sym}(X/A)$. I didn't give the inverse of the isomorphism because it is not needed to construct the sign map.

In the case $A$ is abelian, there is a homomorphism $\tau_0\colon A^{X/A}\rtimes \mathrm{Sym}(X/A)\to A$ given by $(f,\sigma)\mapsto \prod_{e\in X/A}f(e)$ (sometimes called the transfer map). This gives a homomorphism $\tau\colon \mathrm{Aut}_A(X)\to A$ which is independent of the isomorphism constructed above since isomorphisms coming from different sections (i.e. transversals) differ by an inner automorphism and $A$ is abelian.

Now consider the group $A=C_2$ (the two element cyclic group) and consider the set $E$ of distinct pairs of elements from $[n]=\{1,\ldots, n\}$. Then $C_2$ acts freely on the right of $E$ by having the nontrivial element switching the coordinates and the action of $S_n$ on $E$ commutes with this. We then the composed homomorphism $S_n\to \mathrm{Aut}_{C_2}(E)\xrightarrow{\,\tau\,} C_2$.

As usual we just then need to check the transposition $(1,2)$ is sent to the nontrivial element of $C_2$, but this follows easily if you choose as your transversal the ordered pairs $(i,j)$ with $i<j$ and follow through the construction.

There are already enough answers but here is a rewrite of Poonen's and De Bruyn's answer in the language of wreath products.

Let $A$ be a group acting freely on the right of a finite set $X$. Then $\mathrm{Aut}_A(X)$ is well known to be isomorphic to the permutational wreath product $A\wr \mathrm{Sym}(X/A)=A^{X/A}\rtimes \mathrm{Sym}(X/A)$. The isomorphism depends on choosing a transversal for $X/A$ but any two isomorphisms are conjugate by an element of $A^{X/A}$. In concrete terms, choose a section $t\colon X/A\to X$ and send $f\in \mathrm{Aut}_A(X)$ to $(F,f'})$ where $f'$ is the permutation of $X/A$ induced by $f$ and $F\colon X/A\to A$ is defined on $e\in X/A$ by $f(t(e))=t(f'(e))F(e)$. If $t'\colon X/A\to X$ is another section, then there is $h\in A^{X/A}$ such that $t'(e)=t(e)h(e)$ for all $e\in X/A$ and the isomorphism constructed from $t'$ is conjugate to the one constructed from $t$ by $(h,1)\in A\wr \mathrm{Sym}(X/A)$. I didn't give the inverse of the isomorphism because it is not needed to construct the sign map.

In the case $A$ is abelian, there is a homomorphism $\tau_0\colon A^{X/A}\rtimes \mathrm{Sym}(X/A)\to A$ given by $(f,\sigma)\mapsto \prod_{e\in X/A}f(e)$ (sometimes called the transfer map). This gives a homomorphism $\tau\colon \mathrm{Aut}_A(X)\to A$ which is independent of the isomorphism constructed above since isomorphisms coming from different sections (i.e. transversals) differ by an inner automorphism and $A$ is abelian.

Now consider the group $A=C_2$ (the two element cyclic group) and consider the set $E$ of distinct pairs of elements from $[n]=\{1,\ldots, n\}$. Then $A$ acts freely on the right of $E$ by having the nontrivial element switching the coordinates and the action of $S_n$ on $E$ commutes with this. We then the composed homomorphism $S_n\to \mathrm{Aut}_{C_2}(E)\xrightarrow{\,\tau\,} C_2$.

As usual we just then need to check the transposition $(1,2)$ is sent to the nontrivial element of $C_2$, but this follows easily if you choose as your transversal the ordered pairs $(i,j)$ with $i<j$ and follow through the construction.

There are already enough answers but here is a rewrite of Poonen's and De Bruyn's answer in the language of wreath products.

Let $A$ be a group acting freely on the right of a finite set $X$. Then $\mathrm{Aut}_A(X)$ is well known to be isomorphic to the permutational wreath product $A\wr \mathrm{Sym}(X/A)=A^{X/A}\rtimes \mathrm{Sym}(X/A)$. The isomorphism depends on choosing a transversal for $X/A$ but any two isomorphisms are conjugate by an element of $A^{X/A}$. In concrete terms, choose a section $t\colon X/A\to X$ and send $f\in \mathrm{Aut}_A(X)$ to $(F,f')$ where $f'$ is the permutation of $X/A$ induced by $f$ and $F\colon X/A\to A$ is defined on $e\in X/A$ by $f(t(e))=t(f'(e))F(e)$. If $t'\colon X/A\to X$ is another section, then there is $h\in A^{X/A}$ such that $t'(e)=t(e)h(e)$ for all $e\in X/A$ and the isomorphism constructed from $t'$ is conjugate to the one constructed from $t$ by $(h,1)\in A\wr \mathrm{Sym}(X/A)$. I didn't give the inverse of the isomorphism because it is not needed to construct the sign map.

In the case $A$ is abelian, there is a homomorphism $\tau_0\colon A^{X/A}\rtimes \mathrm{Sym}(X/A)\to A$ given by $(f,\sigma)\mapsto \prod_{e\in X/A}f(e)$ (sometimes called the transfer map). This gives a homomorphism $\tau\colon \mathrm{Aut}_A(X)\to A$ which is independent of the isomorphism constructed above since isomorphisms coming from different sections (i.e. transversals) differ by an inner automorphism and $A$ is abelian.

Now consider the group $A=C_2$ (the two element cyclic group) and consider the set $E$ of distinct pairs of elements from $[n]=\{1,\ldots, n\}$. Then $A$ acts freely on the right of $E$ by having the nontrivial element switching the coordinates and the action of $S_n$ on $E$ commutes with this. We then the composed homomorphism $S_n\to \mathrm{Aut}_{C_2}(E)\xrightarrow{\,\tau\,} C_2$.

As usual we just then need to check the transposition $(1,2)$ is sent to the nontrivial element of $C_2$, but this follows easily if you choose as your transversal the ordered pairs $(i,j)$ with $i<j$ and follow through the construction.

There are already enough answers but here is a rewrite of Poonen's and De Bruyn's answer in the language of wreath products.

Let $A$ be a group acting freely on the right of a finite set $X$. Then $\mathrm{Aut}_A(X)$ is well known to be isomorphic to the permutational wreath product $A\wr \mathrm{Sym}(X/A)=A^{X/A}\rtimes \mathrm{Sym}(X/A)$. The isomorphism depends on choosing a transversal for $X/A$ but any two isomorphisms are conjugate by an element of $A^{X/A}$. In concrete terms, choose a section $t\colon X/A\to X$ and send $f\in \mathrm{Aut}_A(X)$ to $(F,\overline f)$ where $\overline f$ is the permutation of $X/A$ induced by $f$ and $F\colon X/A\to A$ is defined on $e\in X/A$ by $f(t(e))=t(\overline f(e))F(e)$. If $t'\colon X/A\to X$ is another section, then there is $h\in A^{X/A}$ such that $t'(e)=t(e)h(e)$ for all $e\in X/A$ and the isomorphism constructed from $t'$ is conjugate to the one constructed from $t$ by $(h,1)\in A\wr \mathrm{Sym}(X/A)$. I didn't give the inverse of the isomorphism because it is not needed to construct the sign map.

In the case $A$ is abelian, there is a homomorphism $\tau_0\colon A^{X/A}\rtimes \mathrm{Sym}(X/A)\to A$ given by $(f,\sigma)\mapsto \prod_{e\in X/A}f(e)$ (sometimes called the transfer map). This gives a homomorphism $\tau\colon \mathrm{Aut}_A(X)\to A$ whichis independent of the isomorphism constructed above since isomorphisms coming from different sections (i.e. transversals) differ by an inner automorphism and $A$ is abelian.

Now consider the group $A=C_2$ (the two element cyclic group) and consider the set $E$ of distinct pairs of elements from $[n]=\{1,\ldots, n\}$. Then $A$ acts freely on the right of $E$ by having the nontrivial element switching the coordinates and the action of $S_n$ on $E$ commutes with this. We then the composed homomorphism $S_n\to \mathrm{Aut}_{C_2}(E)\xrightarrow{\,\tau\,} C_2$.

As usual we just then need to check the transposition $(1,2)$ is sent to the nontrivial element of $C_2$, but this follows easily if you choose as your transversal the ordered pairs $(i,j)$ with $i<j$ and follow through the construction.

There are already enough answers but here is a rewrite of Poonen's and De Bruyn's answer in the language of wreath products.

Let $A$ be a group acting freely on the right of a finite set $X$. Then $\mathrm{Aut}_A(X)$ is well known to be isomorphic to the permutational wreath product $A\wr \mathrm{Sym}(X/A)=A^{X/A}\rtimes \mathrm{Sym}(X/A)$. The isomorphism depends on choosing a transversal for $X/A$ but any two isomorphisms are conjugate by an element of $A^{X/A}$. In concrete terms, choose a section $t\colon X/A\to X$ and send $f\in \mathrm{Aut}_A(X)$ to $(F,f'})$ where $f'$ is the permutation of $X/A$ induced by $f$ and $F\colon X/A\to A$ is defined on $e\in X/A$ by $f(t(e))=t(f'(e))F(e)$. If $t'\colon X/A\to X$ is another section, then there is $h\in A^{X/A}$ such that $t'(e)=t(e)h(e)$ for all $e\in X/A$ and the isomorphism constructed from $t'$ is conjugate to the one constructed from $t$ by $(h,1)\in A\wr \mathrm{Sym}(X/A)$. I didn't give the inverse of the isomorphism because it is not needed to construct the sign map.

In the case $A$ is abelian, there is a homomorphism $\tau_0\colon A^{X/A}\rtimes \mathrm{Sym}(X/A)\to A$ given by $(f,\sigma)\mapsto \prod_{e\in X/A}f(e)$ (sometimes called the transfer map). This gives a homomorphism $\tau\colon \mathrm{Aut}_A(X)\to A$ which is independent of the isomorphism constructed above since isomorphisms coming from different sections (i.e. transversals) differ by an inner automorphism and $A$ is abelian.

Now consider the group $A=C_2$ (the two element cyclic group) and consider the set $E$ of distinct pairs of elements from $[n]=\{1,\ldots, n\}$. Then $A$ acts freely on the right of $E$ by having the nontrivial element switching the coordinates and the action of $S_n$ on $E$ commutes with this. We then the composed homomorphism $S_n\to \mathrm{Aut}_{C_2}(E)\xrightarrow{\,\tau\,} C_2$.

As usual we just then need to check the transposition $(1,2)$ is sent to the nontrivial element of $C_2$, but this follows easily if you choose as your transversal the ordered pairs $(i,j)$ with $i<j$ and follow through the construction.