I'm trying to solve the following problem. I have a solution, I'm tripping over myself explaining part of it. Is my solution clear? (Particularly part 2) If not, can someone help me clarify it? Thanks!
Problem:
If a group $G$ is isomorphic to $H$, prove that $\text{Aut}(G)$ is isomorphic to $\text{Aut}(H)$
Solution:
Let $\mathcal{G} \in \text{Aut}(G)$ and $\phi:G \rightarrow H$. By assumption $\phi$ is an isomorphism. Define: $$\Phi:\text{Aut}(G) \rightarrow \text{Aut}(H) \\ \mathcal{G} \mapsto \phi\mathcal{G}\phi^{-1}$$We want to show $\Phi$ is an isomorphism.
Proceed in 3 steps: Show $\Phi$ injects, show $\Phi$ surjects, show $\Phi$ is a homomorphism.
Suppose $\Phi(\mathcal{G})=\Phi(\mathcal{G'})$ $$\Phi(\mathcal{G})=\Phi(\mathcal{G'}) \\ \implies \phi\mathcal{G}\phi^{-1} = \phi\mathcal{G}'\phi^{-1} \\ \implies \mathcal{G} = \mathcal{G'} \text{ [by taking }\phi^{-1} \text{of both sides, followed by }\phi \text{]}$$
Suppose we have $\mathcal{H} \in \text{Aut}(H)$. We want to show there exists $\mathcal{G}$ such that $\Phi(\mathcal{G})=\mathcal{H}$. Construct such a $\mathcal{G}$ by letting $\mathcal{G} = \phi^{-1}\mathcal{H}\phi$. To check $\phi^{-1}\mathcal{H}\phi \in \text{Aut}(G)$ we need to check $\phi^{-1}\mathcal{H}\phi$ is a bijection on $G$ and that its a homomorphism on $G$. Bijection of $\phi^{-1}\mathcal{H}\phi$ follows from the composition of 3 bijections. We check homomorphism by considering the action on 2 arbitrary elements of $G$ say $g_1$ and $g_2$. $$\phi^{-1}\mathcal{H}\phi(g_1 g_2)=\phi^{-1}\mathcal{H}[\phi(g_1)\phi(g_2)] = \phi^{-1}[\mathcal{H}\phi(g_1)\mathcal{H}\phi(g_2)]=\phi^{-1}\mathcal{H}\phi(g_1)\phi^{-1}\mathcal{H}\phi(g_2)$$ By homomorphic properties of $\phi$, $\phi^{-1}$, and $\mathcal{H}$. Thus $\phi^{-1}\mathcal{H}\phi \in \text{Aut}(G)$. We just need to check that the element of $\text{Aut}(G)$ so defined will send us to the given $\mathcal{H}$. $$\Phi(\phi^{-1}\mathcal{H}\phi) = \phi\phi^{-1}\mathcal{H}\phi\phi^{-1} = \mathcal{H}$$
Let $\mathcal{G_1}$,$\mathcal{G_1} \in \text{Aut}(G)$. Then: $$\Phi(\mathcal{G_1}\mathcal{G_1})=\phi\mathcal{G_1}\mathcal{G_1}\phi^{-1}=\phi\mathcal{G_1}\phi^{-1}\phi\mathcal{G_1}\phi^{-1}=\Phi(\mathcal{G_1})\Phi(\mathcal{G_2})$$
It's correct, apart from one particular. It can also be simplified.
First, prove that, for $\alpha\in\operatorname{Aut}(G)$, $\phi\alpha\phi^{-1}\in\operatorname{Aut}(H)$: this is essentially obvious (but is missing from your proof). Thus $\Phi$ is well defined.
Second, prove that $\Psi\colon\operatorname{Aut}(H)\to\operatorname{Aut}(G)$ defined by $\Psi(\delta)=\phi^{-1}\delta\phi$ is the two-sided inverse of $\Phi$. Note there's no need to check $\Psi$ is well defined, because it's the same argument as above with $G$ and $H$ swapped. Thus $\Phi$ is bijective.
Third, prove $\Phi$ is a homomorphism (your check is right).