If $G$ is isomorphic to $H$, show ${\rm Aut}(G)$ is isomorphic to ${\rm Aut}(H)$

Question

For every $\alpha\in{\rm Aut}(G)$, I've defined $A:H\rightarrow H$ by $$A(h)=\phi(\alpha(\phi^{-1}(h)))$$ where $\phi$ is an isomorphism from G onto H, and I've shown that $A\in {\rm Aut}(H)$. What I can't figure out is how to show that every $B\in {\rm Aut}(H)$ can be expressed as $\phi\beta\phi^{-1}$ for some $\beta\in{\rm Aut}(G)$ or how to show that this mapping is an isomorphism. How can I go about this?

Answer 1

Hint : if $B \in \text{Aut}(H)$, then you can consider $ \beta = \phi^{-1} B \phi \in \text{Aut}(G)$.

Answer 2

You have shown that the map $\Phi \colon \operatorname{Aut}(G) \to \operatorname{Aut}(H)$ with $\Phi(\alpha) = \phi \alpha \phi^{-1}$ is well-defined. This is a group homomorphisms because $$ \Phi(\alpha \alpha') = \phi (\alpha \alpha') \phi^{-1} = \phi \alpha \phi^{-1} \phi \alpha' \phi^{-1} = \Phi(\alpha) \Phi(\alpha'). $$ In the same way we get the group homomorphism $$ \Psi \colon \operatorname{Aut}(H) \to \operatorname{Aut}(G) \quad\text{with}\quad \Psi(\beta) = \phi^{-1} \beta \phi. $$ Then $\Phi(\Psi(\beta)) = \beta$ and $\Psi(\Phi(\alpha)) = \alpha$, so $\Phi$ is already a group isomorphism with $\Phi^{-1} = \Psi.$