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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 $A$.

You are assuming that $A$ is an abelian normal subgroup of a group $G$ of odd order, with the property that all automorphisms $\alpha$ of $G$ that induce the identity on $G/A$ act trivially on $A$. You want to prove that all endomorphisms of $G$ that induce the identity on $A$ are automorphisms.

I will prove the contrapositive of that statement. Suppose that there exists an endomorphisms $\phi$ of $G$ with nontrivial kernel $K$, where $\phi(A)=A$ and $\phi$ induces the identity on $G/A$. I will prove your assertion by constructing an automorphism of $G$ that induces the identity on $G/A$ but not on $A$.

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 $G/A$, but not on $A$, so this proves your statemen.

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 $\phi$ fixes $A$ and 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 $A$.

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 $A$.

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 $\phi$ fixes $A$ and 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 induces acts trivially on $G/A$. So it is not true that every automorphism of $G$ that acts trivially on $G/A$ acts trivially on $A$.

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 $\phi$ fixes $A$ and 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 $A$.