extending homomorphisms to semidirect products Let $A = F_2$ 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 with $Z(G) = 1$, 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.
EDIT: Changed condition to require that $Z(G) = 1$. If this is false, What is the obstruction to this failing? Can it be measured by some cohomology class?
 A: Edit : a counter-example $Z(G)$ is trivial but images are not conjugate with each other. Take $n\geq 3$ (or even $n=3$), $G=\mathfrak{S}_n$, the symmetric group over $n$ elements. It is well known that $g_1:=(1,2)$ and $g_2:=(1,2,...,n)$ generate $G$. Furthermore by the property of the free group, we have a group morphism from $A$ to $\mathfrak{S}_n$ sending the first generator $\gamma_1$ of $A$ to $g_1$ and the second generator $\gamma_2$ of $A$ to $g_2$. This is our group morphism $f_A$ (it is onto because $g_1,g_2$ generate $G$).
On the other hand take $C:=\mathbb{Z}_2$ (the cyclic group over $2$ elements) and take $\beta:C\rightarrow Aut(A)$ defined by $\beta(0)=Id_A$ and $\beta(1)(\gamma_1):=\gamma_2$ and $\beta(1)(\gamma_2):=\gamma_1$. This last definition makes $\beta(1)$ into a group automorphism of $A$ of order $2$ (it exchanges the two generators). 
I want to show that the $f_A$ does not extend to $A\rtimes C$. Let us simply assume that it does then we can find $g_1\in G$ such that :
$$f_A(\gamma_2)=f_A(\beta(1)(\gamma_1))=g_1f_A(\gamma_1)g_1^{-1}$$
Hence in $G$, we have that $(1,2,...,n)$ is conjugate with $(1,2)$. Clearly this cannot happen for $n\geq 3$. So that we cannot extend $f_A$ to $A\rtimes C$. More generally you see that if such extension exists, $f_A(\beta(c)(a))$ will be conjugate to $f_A(a)$ for all $c\in C$, I claim that this is a very strong condition (unless $C$ acts by inner automorphism in $A$, in which case we can assume that the action is trivial). 
Take now the situation as you give it. We want to find $f$ such that $f_{|A}=f_A$. We define $g_c$ as you do it and then define first $f_0$ by :
$$f_0(a,c):=f_A(a)g_c $$
$$f_0(a,c)f_0(b,d)=f_A(a)g_cf_A(b)g_d=f_A(a)f_A(\beta(c).b)g_cg_d$$
$$f_0(a,c)f_0(b,d)=f_A(a\beta(c).b)g_{cd}(g_{cd}^{-1}g_cg_d) $$
$$f_0(a,c)f_0(b,d)=f_0((a,c)(b,d))z(c,d)$$
Here we defined $z(g,h):=g_{cd}^{-1}g_cg_d$. Since $\beta(c)\beta(d)$ is $\beta(cd)$ one can check that $z(c,d)$ must centralize the image of $f_A$. If $f_A$ happens to be surjective (this is the case here) we deduce that for any $c,d\in C$, $z(c,d)\in Z(G)$. Now $f_0$ as we defined it is a group morphism if and only if we may choose $(g_c)_{c\in C}$ such that $z(c,d)=1_G$ for all $c,d$. 
Remark that :
$$z(c,d)+z(cd,e)=z(d,e)+z(c,de) $$
This shows that $z\in Z^2(C,Z(G))$ ($Z(G)$ is considered as a trivial $C$-module). 
Remark that $(g_c')$ does the same action as $(g_c)$ if and only if for all $c\in C$, $g_c'=t_cg_c$ with $t_c\in Z(G)$. Hence if we define $z'$ associated to $(g_c')$ we have that :
$$z'(c,d)=z(c,d)+t_c+t_d-t_{cd} $$
Here this is the additive notation in $Z(G)$. Clearly, we said that $[z]\in H^2(C,Z(G))$ (considering $Z(G)$ as a trivial $C$-module) does not depend on the particular choice of $(g_c)$ (provided that it does the prescribed action). Hence the proposition is the following :

Given a prescribed action of $C$ on $G$ (by giving one family $(g_c)$) through $f_A$ then one may extend $f_A$ to $A\rtimes C$ if and only if $[z]\in H^2(C,Z(G))$ is null. 

This is the obstruction you are looking for. We remark that if $Z(G)$ is trivial then there is no obstruction. 
