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. 



Automorphisms and epimorphisms of finite groups
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.


