TITLE: Automorphisms and epimorphisms of finite groups
QUESTION [9 upvotes]: All groups in this question are finite, and epimorphism means surjective group homomorphism.
Suppose I have two epimorphisms $f,g\colon G\to H$.  This implies that $\ker(f)$ and $\ker(g)$ have the same composition factors, but they need not be isomorphic. I'll say that $f$ and $g$ are compatible if $g=fh$ for some automorphism $h$ of $G$.  This would imply that $\ker(g)\simeq\ker(f)$, so it is not always true.  I ask: does there always exist $K$ and an epimorphism $p\colon K\to G$ such that $fp$ and $gp$ are compatible?
If $G$ is nilpotent we can reduce to the case where it is a $p$-group, then I think we can take $K$ to be the initial example of an $k$-generator group of exponent $p^n$ and nilpotence class $c$, for sufficiently large $k$, $n$ and $c$.  In particular, if $G$ is an abelian $p$-group I think we can take $K=C_{p^n}^k$ for sufficiently large $k$ and $n$.  But I am not sure what to do when $G$ is not nilpotent.

REPLY [13 votes]: You could start with $K$ being the free group with the elements of $G$ as its free basis, with  $p$ the obvious map onto $G$. Then the required automorphism $h$ of $K$ to make the two composite maps compatible is just a permutation of the free basis elements of $K$.
Of course this $K$ is infinite, and you are looking for a finite group. But to get that, we could replace $K$ by $K/N$, for any characteristic  subgroup $N$ of  finite index in $K$, and with $N \le \ker p$. You could take $N$ to be the intersection of the kernels of all epimorphisms from $K$ to $G$, of which there are finitely many.