Let $A = F_2$ be the free group on two generators (not sure if this is important.)

Suppose you have a semidirect product $A\rtimes C$ coming from some homomorphism $\beta: C\rightarrow \text{Aut}(A)$.

Let $G$ be a finite group, and let $f_A: A\twoheadrightarrow G$ be a surjection such that $$f_A(^c\cdot) := f_A\circ\beta(c)\equiv f_A\mod\text{Inn}(G)\qquad\text{for all $c\in C$}$$

then, under these conditions, must there exist:

- A homomorphism $f : A\rtimes C\twoheadrightarrow G$ such that $f|_A = f_A$, or equivalently...
- A homomorphism $f_C : C\rightarrow G$ such that we have $$(f_A\circ\beta(c))(a) := f_A(^ca) = f_C(c)f_A(a)f_C(c)^{-1}\qquad\text{for all $c\in C, a\in A$}$$

*Remark:* Our assumptions tell us that for every $c\in C$, there is a $g_c\in G$ such that $f_A(^ca) = g_cf_A(a)g_c^{-1}$ for all $a\in A$, so what I'm asking is if it's always possible to pick the $g_c$'s wisely so that the map $c\mapsto g_c$ is a homomorphism.

This question came from me thinking about Teichmuller level structures and how they classify torsors, so in my context $1\rightarrow A\rightarrow A\rtimes C\twoheadrightarrow C\rightarrow 1$ is an exact sequence of fundamental groups associated to some punctured curve (either $\mathbb{P}^1$ minus 3 points or an elliptic curve punctured once).

*EDIT:* As Dave Witte Morris points out, this is not possible in general. Though, I wonder: Are there conditions we can put on $G$ or $f_A$ to ensure that the statement *is* true? If $G$ is abelian, then $f_C$ could literally be anything, but that's quite a strong condition. Can we say anything for nonabelian $G$? I'm trying to understand exactly why this fails... Is there some kind of cohomology going on?