Let $F$ be a finite group, let $\pi_g$ be the fundamental group of the closed oriented genus $g$ surface $S_g$ and let $Mod_g$ denote the mapping class group of $S_g$. Consider the action of $Mod_g$ on the "character variety'' $R(\pi_g,F)=Hom(\pi_g, F)/Aut(F)$ by precompositions.

The question is "how transitive is the action of $Mod_g$ on $R(\pi_g,F)$?" It is clear (see e.g. Daniele's answer) that $Mod_g$ cannot send an epimorphism to a non-epimorphism. Thus, consider the subset $E(\pi_g,F)\subset R(\pi_g, F)$ consisting of equivalence classes of epimorphisms. There is one more invariant that $Mod_g$ has to preserve, namely, every homomorphism $f: \pi_g\to F$ represents an element $c_f$ of $H_2(F, {\mathbb Z})$, which is the image of the fundamental class of $S_g$ under the induced map
$$H_2(f): H_2(S_g)\to H_2(K(F,1)).$$
Thus, we get a map $c: R(\pi_g,F)\to Q_F:=H_2(F, {\mathbb Z})/Aut(F), f\mapsto c_f$. The classes $c_f$ (modulo $Aut(F)$) are, clearly, preserved by the action of $Mod_g$, so $Mod_g$ preserves each fiber of the map $c$.

Amazingly, it turns out that for every simple nonabelian group $Q$, "stably" (i.e., for fixed $F$ and all sufficiently large genera $g$), the map $c$ completely classifies the $Mod_g$-orbits on $E(\pi_g,F)$:

For every simple nonabelian $Q$, if $g$ is sufficiently large, two epimorphisms $f_1, f_2$ belong to the same orbit $\iff$ $c_{f_1}=c_{f_2}$. This is proven in Theorem 1.3 of the paper by N.Dunfield and W.Thurston "Finite covers of random 3-manifolds". Furthermore, for all large $g$, the action of $Mod_g$ on every orbit $O$ is via the full alternating group of $O$. What happens for small $g$'s is very interesting but unclear.

A similar result, based on the previous work of Convey and Parker, in the context of the action of the braid group, was proven in: M. D. Fried and H. Volklein, "The inverse Galois problem and rational points on moduli spaces," Math. Ann. 290 (1991), 771–800, see footnote on page 5 of the paper by Dunfield and Thurston.

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