Let $G$ be finite of odd order, and let $A$ be an abelian normal subgroup of $G$. Suppose that $\phi$ is an endomorphism of $G$ with nontrivial kernel $K$, where $\phi(A)=A$ and $\phi$ induces the identity on $G/A$. Then clearly $K \le A$. Let $H = {\rm im}(\phi)$.

If $H \cap K \ne 1$, then $\phi^2$ still induces the identity on $G/A$ and has smaller image than $\phi$. So, by replacing $\phi$ by a power, we may assume that $H \cap K = 1$ and hence $H$ is a complement to $K$ in $G$.

Now we can define an automorphism of $G$ that fixes $H$ and $K$, induces the identity on $H$ and inverts every element of $K$. Since $K \le A$, this automorphism acts trivially on $G/A$. So it is not true that every automorphism of $G$ that acts trivially on $G/A$ acts trivially on $A$.