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This question already has an answer here:

Say $G \cong H$ are isomorphic groups. Show $Aut(G) \cong Aut(H)$

I just made this up so I'm not sure if actually $Aut(G) \cong Aut(H)$ is true but I'm $99.9\%$ sure this should be true

I'm having trouble with the proof :'(

Let $\theta:G \rightarrow H$ be the isomorphism and $\phi\in Aut(G)$

Let $g_1,g_2\in G$ and let $h_1,h_2 \in H \ \ with \ \ h_1 = \theta(g_1) \ \ and \ \ h_2 = \theta(g_2)$

If $\phi(g_1) = g_2$ I want to show that $\exists \alpha\in Aut(H)$ such that

$\alpha(h_1) = h_2$

I tried composing $\theta$ and $\phi$ to get into $H$ but that doesn't get me anywhere. Is there something wrong with this approach?

Thanks ! :D

Wait a minute....

$\theta(\phi(g_1)) = \theta(g_2) = h_2$

$\theta(\phi(g_2)) = \theta(g_1) = h_1$

I think I see something here, $h_1$ and $h_2$ were permuted in the same way $g_1$ and $g_2$ were but $\theta \phi : G\rightarrow H$ so it cant be in $Aut(H)$

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marked as duplicate by Watson, Martin Sleziak, C. Falcon, Stahl, Shailesh Dec 2 '16 at 4:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You need to show that given an isomorphism $f:G \to H$, for each automorphism on $G$ you can find a unique isomorphism on $H$ (thus giving a $1-1$ mapping). For this just use the isomorphism to take the automorphism on $G$ and define an automorphism on $H$ by mapping elements (and confirm this actually gives an automorphism on $H$). Then uniqueness of the elements mapped will follow automatically, and you have a $1-!$ mapping in the other direction given by considering the inverse map $f^{-1}$. Then you can either prove that the two mappings between automorphisms are inverse to each other, or just appeal to the theorem that two sets are in $1-1$ correspondance if you can find an injective mapping from each set to the other.

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Let $\theta : G \to H$ be an isomorphism. Define a map $\Phi : \operatorname{Aut}(G) \to \operatorname{Aut}(H)$ by $\Phi(f) = \theta \circ f \circ \theta^{-1}$. (Show that $\Phi(f)$ is indeed an automorphism of $H$.)

The map $\Phi$ is easily seen to be a homomorphism. It has the inverse $\Psi(g) = \theta^{-1} \circ g \circ \theta$. It follows that $\Phi$ is an isomorphism.

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What you are trying to prove is true, but you can't focus on just two elements of each group, you need to look at the whole groups.

A composition is a good idea, but $\theta\circ \phi$ goes from $G$ to $H$, whereas you're looking for a map from $H$ to $H$. What can you do with a map from $G$ to turn it into a map from $H$?

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