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Let graph $G$ be isomorphic with $H$. I would like to show $\operatorname{Aut}(G)=\operatorname{Aut}(H)$, where $\operatorname{Aut}(G)$=Set of automorphisms of graph $G$).

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    The *sets* of automorphisms are hardly ever the *same*. But something stronger is true. The automorphism *groups* are isomorphic. My feeling is that the assertion does not even require proof. But if one wants a proof, it is automatic.2011-12-22

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HINT: If $G$ is isomorphic with $H$, there exists an isomorphism $\varphi$ from $G$ to $H$. Now, let $h \in \operatorname{Aut}(G)$, then $\varphi\circ h \circ \varphi^{-1} \in \operatorname{Aut}(H)$. Also, for each $h$ this automorphism is unique, thus $|\operatorname{Aut}(G)| \leq |\operatorname{Aut}(H)|$. Can you take it from here?

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    To be honest, i don't think that both groups (unless$H$= G) can be equal to one another, since the domains and codomains of the automorphisms are different, thus the functions cannot be the same. However, they can be isomorphic as you already mentioned.2011-12-22