If $G$ is isomorphic to $H$, then ${\rm Aut}(G)$ is isomorphic to ${\rm Aut}(H)$ 
If a group $G$ is isomorphic to $H$, prove that ${\rm Aut}(G)$ is isomorphic to ${\rm Aut}(H)$.

Can someone provide a step by step solution? Explaining along the way our strategy in proving this.
 A: Let $\phi:G\to H$ be an isomorphism. A useful intuition is that $G$ and $H$ are just two manifestations of the same object, except that the group elements may have different names in the two cases. The mapping $\phi$ tells us for each element $g\in G$ which element $\phi(g)\in H$ corresponds to this element. In other words, $\phi$ tells us how to rename the elements of $G$ to obtain elements of $H$. The inverse of this isomorphism, $\phi^{-1}:H\to G$ tells us, how to get from an element $h\in H$ back to an element $\phi^{-1}(h)\in G$.
Now suppose $\alpha:G\to G$ is an automorphism. (Note that an automorphism may be regarded as "a structure-preserving renaming of the elements within the group".) Since we regard $H$ as just another version of $G$ with different names of the elements, we should expect this automorphism to correspond to some automorphism $\Phi(\alpha):H\to H$. But how to get this automorphism? 
Remembering that we consider an element $h\in H$ to be just a renamed version of some element $g\in G$, namely $g = \phi^{-1}(h)$, we are led to the idea that we may define the desired automorphism $\Phi(\alpha)$ by simply renaming the element $h$ to "whatever it's called in $G$", then doing what $\alpha$ does, to obtain the result, and then renaming it back, to obtain "the name of the result in $H$".
If we trace what this does, we obtain: $$h\mapsto \phi^{-1}(h)\mapsto \alpha(\phi^{-1}(h))\mapsto \phi(\alpha(\phi^{-1}(h))),$$ so the automorphism $\Phi(\alpha): H\to H$ will be defined by
$$\Phi(\alpha)=\phi\circ\alpha\circ\phi^{-1}.$$
You may verify that $\Phi:\operatorname{Aut}(G)\to\operatorname{Aut}(H)$, defined in this way, is indeed an isomorphism of groups. (Note that $$\Phi(\beta)\circ\Phi(\alpha)=(\phi\circ\beta\circ\phi^{-1})\circ(\phi\circ\alpha\circ\phi^{-1})=\phi\circ\beta\circ\alpha\circ\phi^{-1}=\Phi(\beta\circ\alpha).$$)
We can visually express the definition of $\Phi$ by the following commutative diagram:
$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\newcommand{\ua}[1]{\bigg\uparrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
G & \ra{\alpha} & G\\
\da{\phi} & & \ua{\phi^{-1}}\\
H & \ras{\Phi(\alpha)} & H\\
\end{array}$$
A: If $C$ is any category with objects $x,y$ such that $x \cong y$, then $\mathrm{Aut}(x) \cong \mathrm{Aut}(y)$ as groups. Namely, if $f : x \to y$ is an isomorphism, then every automorphism $g : x \cong x$ can be "conjugated" to an automorphism $y \xrightarrow{f^{-1}} x \xrightarrow{g} x \xrightarrow{f} y$, and vice versa.
A: HINT: It should be clear that the first thing to do is name an isomorphism between $G$ and $H$: let $h:G\to H$ be an isomorphism. Now you want to use $h$ somehow to construct an isomorphism between $\operatorname{Aut}(G)$ and $\operatorname{Aut}(H)$. Suppose that $\varphi\in\operatorname{Aut}(G)$. How could you use $\varphi$ and $h$ to construct an automorphism of $H$? There’s only one reasonable thing to try. 
An automorphism of $H$ maps $H$ to $H$, so start with an arbitrary $x\in H$. Use $h^{-1}$ to find the corresponding element $h^{-1}(x)$ of $G$, apply $\varphi$ to that element to get another element of $G$, and use $h$ to find the corresponding element of $H$. This may sound a bit complicated, but it’s just a composition of functions. Then prove that the map that you’ve constructed from $H$ to $H$ really is an automorphism.
To show that every automorphism of $H$ arises in this way, just start with an automorphism $\psi$ of $H$ and use the same ideas to find a $\varphi\in\operatorname{G}$ that gets sent to $\psi$ by the map described in the preceding paragraph.
