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

Properties of Isomorphisms acting on groups:
Suppose that $\phi$ is an isomorphism from a group G onto a group H, then:
 1. $\phi^{-1}$ is an isomorphism from H onto G.
 2. G is Abelian if and only if H is Abelian
 3. G is cyclic if and only if H is cyclic.
 4. If K is a subgroup of G, the $\phi$(K)={$\phi$(k)|k$\in$K} is a subgroup of H.
 5. If L is a subgroup of H, then $\phi^{-1}$L={g$\in$G|$\phi$(g)$\in$L} is a subgroup of G.
 6. $\phi$(Z(G))=Z(H)
Steps to proving G is isomorphic to H:


*

*Define a function $\phi$ from G to H  

*Prove that $\phi$ is one-to-one  

*Prove that $\phi$ is onto  

*Show that $\phi (ab)=\phi (a)\phi (b)$ for all $a,b\in G$


Proof:
Suppose that $\phi:G\rightarrow H$ is an isomorphism.  
Since G is a group, it has elements $a,b\in G$ such that $ab\in G$.$\space \space$H is a group as well.$\space \space$Therefore, $\phi(a)=x\in H$, $\phi(b)=y\in H$, and $\phi(a)\phi(b)=\phi(ab)=xy\in H$.  Thus $\phi$ is one-to-one and operation preserving.
For any element $h\in H$, there is a $g\in G$ such that $\phi(g)=h$, so $\phi$ is onto.
The isomorphism has an inverse operation $\phi^{-1}:H\rightarrow G$ such that $\phi^{-1}(h)=g\in G$.
Similarly, $\psi(a)=r\in G$, $\psi(b)=s\in G$, and $\psi(a)\psi(b)=\psi(ab)=rs\in G$.  Thus $\psi$ is one-to-one and operation preserving.
For any element $\bar g\in G$, there is a $g\in G$ such that $\psi(g)=\bar g$, so $\psi$ is onto.$\space \space$So $\psi$ is an automorphism on G.
The automorphism $\theta(\psi):H\rightarrow H$ is defined by $\theta(\psi)=\phi∘\psi∘\phi^{−1}$ and is an isomorphism.
 A: 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.
A: 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 $\theta$.
