Property of isomorphic normal subgroups

Question

Suppose $A$ and $B$ are two normal subgroups of given $G$ group and $A \cong B$. I'm wondering how to prove that $\phi(A)\cong B$ where $\phi \in \operatorname{Aut}(G)$.

Answer 1

Recall that an automorphism is an isomorphism of $G$. If you restrict the domain of $\phi$ to $A$, it is still bijective with its image; so you have a bijective homomorphism between $A$ and $\phi(A)$. Since $A \cong B$ by assumption, we have $\phi(A) \cong A \cong B$.

Answer 2

Your wording is ambiguous. For instance, perhaps you are wondering "how to prove that there exists $\phi \in \operatorname{Aut}(G)$ such that $\phi(A) \cong B$''. It is quite possible that no such automorphism exists, for instance $G=\mathbb{Z}$, $A=2\mathbb{Z}$, $B=3\mathbb{Z}$.