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?

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.