Let $G$ be a group of odd order. It is known that if every central automophism of $G$ acts trivially on the center, then $G$ is purely non-abelain, this amounts to saying that every central endomorphism $u$ of $G$ (an endomorphism such that $x^{-1}u(x) \in Z(G)$, for all $x \in G$) is an automorphism.

I could generalize this (using some ring theory) to the case of the automorphisms acting trivially on a quotient of an abelain normal subgroup: Let $A$ be an abelian normal subgroup of $G$. If every automorphism of $G$ acting trivially on $G/A$ leaves $A$ elementwise fixed, then every endomorphism of $G$ acting trivially on $G/A$ is an automorphism.

I wonder if one can see a straightforward purely group theoretic proof of this result.

Have you a counter example, if one replace $A$ by any normal subgroup of $G$? More precisely, given a normal subgroup $N$ of $G$ such that any automorphism $u$ acting trivially on $G/N$ (that is $x^{-1}u(x) \in N$, for all $x \in G$), leaves $N$ elementwise fixed. Is it true that every endomorphism $v$ of $G$ such that $x^{-1}v(x) \in N$, for all $x \in G$, is an automorphism?

Thanks in advance.

Let $G$ be a group of odd order. It is known that if every central automophism of $G$ acts trivially on the center, then $G$ is purely non-abelain, this amounts to saying that every central endomorphism $u$ of $G$ (an endomorphism such that $x^{-1}u(x) \in Z(G)$, for all $x \in G$) is an automorphism.

I could generalize this (using some ring theory) to the case of the automorphisms acting trivially on a quotient of an abelain normal subgroup: Let $A$ be an abelian normal subgroup of $G$. If every automorphism of $G$ acting trivially on $G/A$ leaves $A$ elementwise fixed, then every endomorphism of $G$ acting trivially on $G/A$ is an automorphism.

I wonder if one can see a straightforward purely group theoretic proof of this result.

Thanks in advance.