Finding isomorphism between two automorphism groups. I am trying to prove "If $G\cong H$, then $Aut(G) \cong Aut(H)$". I've constructed a homomorphism that seemed "natural", but it turns out to be not injective. Is there a methodical way of finding this isomorphism?
 A: Since $G$ is isomorphic with $H$, there is an isomorphism $\Phi : G \rightarrow H$, so $H$ can be thought of as $\Phi (G)$ and likewise $G$ can be thought of as $\Phi^{-1}(H)$. So one may define $\Psi : Aut(G) \rightarrow Aut(H)$ by $\Psi (\phi) = \Phi \circ \phi \circ \Phi^{-1}$. The idea is that to get a corresponding automorphism on $H$ from one on $G$, you translate back in forth via $\Phi$ (and its inverse). It is immediate to check that $\Psi$ is well-defined and bijective. To show its a homomorphism, you do the usual trick 
$\Psi \circ (\phi \circ \psi) \circ \Psi^{-1} = (\Psi \circ \phi \circ \Phi^{-1}) \circ (\Phi \circ  \psi \circ \Psi^{-1})$
Similarly $\Psi$ preserves inverses and immediately it preserves the unit, and hence is an isomorphism. 
A: Choose an isomorphism $\alpha: G \to H$.  Define a function $\beta: \mathrm{Aut}(G) \to \mathrm{Aut}(H)$ by $\beta(f) = \alpha \circ f \circ \alpha^{-1}$.  Certainly $\beta(f)$ is an automorphism of $H$, then you check that $\beta$ itself is an isomorphism with inverse $g \mapsto \alpha^{-1} \circ g \circ \alpha$.  
A: Fix an isomorphism $\psi\colon G\to H$.  Then we have induced homomorphisms
$\psi^{-1}\circ(-)\circ\psi\colon\operatorname{Aut}H\to\operatorname{Aut}G$ and $\psi\circ(-)\circ\psi^{-1}\colon\operatorname{Aut}G\to\operatorname{Aut}H$...
A: Let $\phi$ be the automorphism $G \to H$.
Send $x \in Aut(H)$ to $\psi(x)=\phi^{-1} x \phi$. This results in a homomorphism going from $G \to G$.
$\psi(x)\psi(y)=\phi^{-1} x \phi \phi^{-1} y \phi=\phi^{-1} xy \phi = \psi (xy)$
$\phi^{-1} x^{-1} \phi = \psi(x^{-1})=\psi(x)^{-1}$ provides it's inverse.
So you can convince yourself this gives $Aut(H) \to Aut(G)$ homomorphism.
For the other direction swap the names $G,H$ and replace $\phi \to \phi^{-1}$. Call that $\Psi$. You see a pair of homomorphisms in both directions such that $\Psi \psi =id_{Aut(H)}$ and $\psi \Psi = id_{Aut(G)}$
