One small observation: if $q$ is sufficiently large compared with the genus of $X$ (at least $2$), then there will be no such curves. Indeed, the size of the automorphism group of a curve is bounded above in terms of its genus, and the number of $\mathbb F_q$-points on a curve is bounded below by the Weil bound. So if $q>>g$, then there aren't enough automorphisms to act transitively.

I suspect that this is quite a hard question to give a sensible answer to. More or less, you are asking to characterise all the curves $X$ over $\mathbb F_q$ such that $Aut(X)$ acts transitively on the points of $X$ with residue field $q^n$. It doesn't seem to me that there will be a nice characterisation of these, unless you restrict to a particular genus.

You can obtain a vector bundle of Lie algebras by taking any locally constant sheaf of groups and applying any of several different constructions that turn groups into Lie algebras. For instance, if you take the associated graded of the descending central series of a group $G$, you get a Lie ring (with bracket coming from the group commutator). So take a locally trivial fibration $X \to S$ with a splitting $f: S \to X$. The fundamental groups $\pi_1(X_s,f(s))$ vary in a local system, and so too do the associated Lie algebras. Then apply Riemann--Hilbert to turn it into a vector bundle.

I concur with your comments -- it does seem like the words "finitely presented" should be included in the definition, even if perhaps they are implied (very non-obviously!) by the other conditions. Thanks for the comments, this has really cleared this up for me!

A small variant on the strategy @WillSawin proposes: if $f$ is a modular function of weight $0$, holomorphic away from the cusp and with a simple pole at the cusp, then there is some complex number $a$ such that $f-aj$ is holomorphic at the cusp (cancel out the leading term of the pole). The same Liouville argument shows that $f-aj$ is constant, i.e. $f=aj+b$. If you want your normalisation conditions $f(e^{2\pi i/3})=0$ and $f(i)=1$, then this forces $a=1$ and $b=0$, so $f=j$.

@AchimKrause By way of analogy, consider the "free abelian group" functor, from {sets} to {abelian groups}. The free abelian group $F_{ab}(S)$ on a set $S$ is characterised by the fact that $\mathrm{Hom}(F_{ab}(S),G)\to\mathrm{Map}(S,G)$ is an isomorphism for all abelian groups $G$. The free group $F(S)$ on $S$ also satisfies this same universal property, but this does not imply that $F(S)=F_{ab}(S)$ in general, exactly because $F(S)$ can be non-abelian. I could envisage something similar happening with strict vs. non-strict Picard groupoids.

@AchimKrause I'm not sure I follow the logic that goes from "the free Picard groupoid on $S$ is not strict" to "there is no free strict Picard groupoid on $S$". That is, because I'm requiring the universal property only for $\mathcal G$ strict, it is not automatic that a free strict Picard groupoid would be a free Picard groupoid, or vice versa. (I wasn't clear in the question that $\mathcal G$ should be assumed strict; I'll clarify that now). Put another way, I believe that $\Omega^\infty\Sigma^2H\mathbb Z$ is not strict, so doesn't give a counterexample to $\mathbb Z$ being free strict.

@AchimKrause By a Picard groupoid, I mean a symmetric monoidal 1-category in which all morphisms are invertible and the functor $X\otimes(-)\colon \mathcal G\to\mathcal G$ is an equivalence for all objects $X$. By a strict Picard groupoid, I mean a Picard groupoid in which the commutativity constraint $\beta_{X,X}\colon X\otimes X \to X\otimes X$ is the identity for all objects $X$. These are sometimes called strictly commutative Picard groupoids.