Equivalance of surjections from a surface group to a free group

Question

Let $g \geq 2$. Let $S = \langle a_1,b_2,...,a_g,b_g | [a_1,b_1] \cdots [a_g,b_g] \rangle$ be the fundamental group of a genus $g$ surface and let $F_g$ be a free group with $g$ generators. Given two surjections $f_1,f_2 : S \to F_g$ is there a way to determine if there are automophisms $\phi: S \to S$ and $\psi: F_g \to F_g$ so that $f_1 = \phi \circ f_2 \circ \psi$?

Answer 1

Reduction of problem, but not an answer:

Suppose $N=\ker f_1 = \ker f_2$. Then $f_1$ and $f_2$ both induce isomorphisms, $f_1' : S/N\to F_g$ and $f_2' : S/N\to F_g$. Then $\phi =f_1'f_2^{\prime -1}$ is an automorphism of $F_g$, and $f_1=\phi f_2$. Thus we only need to find an automorphism $\psi$ of $S$ taking $\ker f_1$ to $\ker f_2$, since then we'll have $\ker f_2\psi = \ker f_1$. Thus the desired result is true if and only if the automorphism group of $S$ acts transitively on normal subgroups $N$ of $S$ with $S/N \cong F_g$.

I'm not sure how to prove that the automorphism group either does or does not act transitively on such subgroups though.