Is the symmetry of a system self-adjoint?

The fact that an operator $A$ commutes with a Hamiltonian $H$ does not necessarily imply that it is self-joint. However, one can without loss of generality assume that symmetries are self-adjoint operators. Suppose that non-self-adjoint operator $A$ is a symmetry, i.e., $A$ satisfies \begin{align} AH=HA \end{align} This condition is equivalent to $H$ commuting with two self-adjoint operators: \begin{align} O_1=A+A^\dagger, \; \; O_2=i(A-A^\dagger), \end{align} i.e., both can be regarded as symmetries. To see this, take the adjoint of both sides of $AH=HA$ and use the fact that $H^\dagger =H$: \begin{align} (AH)^\dagger =(HA)^\dagger \Rightarrow H^\dagger A^\dagger = A^\dagger H^\dagger \Rightarrow H A^\dagger = A^\dagger H \end{align} Hence, $H$ commutes with both $A$ and $A^\dagger$. It follows that $H$ also commutes with any linear combination of these operators, in particular with $O_1=(A+A^\dagger)$ and $O_2=i(A-A^\dagger)$.

Conversely, if $H$ commutes with $O_1$ and $O_2$, then it also commutes with any linear combination $\alpha O_1 + \beta O_2 $. Choosing $\alpha=\tfrac{1}{2}$ and $\beta=\tfrac{-i}{2}$, we recover that $H$ commutes also with $\tfrac{1}{2} O_1 - \tfrac{i}{2} O_2=A$.

Thus, the commutation relation $[A,H]=0$ is completely equivalent to $[O_1,H]=[O_2,H]=0$, showing that one can restrict attention to self-adjoint symmetries without loss of generality.