If a group G is isomorphic to H, prove that Aut(G) is isomorphic to Aut(H)

Suppose $\phi:G\rightarrow H$ is an isomorphism of groups.

Consider $\theta:\text{Aut}(G)\rightarrow\text{Aut}(H)$ defined as follows: let $\psi\in\text{Aut}(G)$, then $\theta(\psi)=\phi\circ\psi\circ\phi^{-1}$.

Now, you must show the following:

$1$) $\theta(\psi)\in\text{Aut}(H)$. This is easy because composition of isomorphisms is an isomorphism and the domains and codomains match.

$2$) $\theta$ is a homomorphism. Multiplication in the group of automorphisms is via composition of functions. Then, let $\psi_1,\psi_2\in\text{Aut}(G)$. Then $\theta(\psi_1\circ\psi_2)=\phi\circ\psi_1\circ\psi_2\circ\phi^{-1}=\phi\circ\psi_1\circ\phi^{-1}\circ\phi\circ\psi_2\circ\phi^{-1}=\theta(\psi_1)\circ\theta(\psi_2)$ since $\psi^{-1}\circ\psi$ is the identity.

$3$) $\theta$ is bijective. In this case, it is easiest to define an inverse map, $\theta^{-1}:\text{Aut}(H)\rightarrow\text{Aut}(G)$. I'll leave this to you, but the definition is similar to the definition of $\theta$.

You are trying to prove that $\phi: G \to H$ and $\psi: G \to G$ are isomorphisms, while you are actually assuming this. Your last line is what you actually need to prove, i.e. that $$ \begin{align} \theta : \text{Aut}(G) &\to \text{Aut}(H) \\ \psi &\mapsto \phi \circ \psi \circ \phi^{-1} \end{align} $$ is an isomorphism of groups. To do this we first need to prove that $\theta$ is well defined, i.e. that its image is indeed contained in $\text{Aut}(H)$, but this is clearly true because $\theta(\psi): H \to H$ is a composition of invertible group morphisms.

Instead of explicitly proving injectivity and surjectivity, though, you could just show that $$ \begin{align} \gamma : \text{Aut}(H) &\to \text{Aut}(G) \\ \varphi &\mapsto \phi^{-1} \circ \varphi \circ \phi \end{align} $$ is an inverse for $\theta$, i.e. that $\gamma \circ \theta = \text{id}_{\text{Aut}(G))}$ and $\theta \circ \gamma = \text{id}_{\text{Aut}(H))}$. Then all you need to prove is that $\theta(\psi_1 \circ \psi_2) = \theta(\psi_1) \circ \theta(\psi_2)$, but this is almost immediate.