# Graph isomorphic

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. – André Nicolas Dec 22 '11 at 23:29
babgen, why don't you accept Dimitri's answer if you think it is so great? – Graphth Dec 30 '11 at 18:47

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?
if $h \in Aut(G)$ and $\varphi^{-1} \in Aut(H)$implies $\varphi\circ h \circ \varphi^{-1} \in Aut(H)$why? – Babak Miraftab Dec 22 '11 at 20:22
$\varphi^{-1}$ is not a member of $Aut(H)$, however, $\varphi^{-1}$ is an isomorphism from $H$ to $G$ and composition of two isomorphisms is again an isomorphism, so $\varphi\circ h \circ \varphi^{-1}$ is an isomorphism. Do you see that $\varphi\circ h \circ \varphi^{-1}$ is a function from $H$ to $H$? – sxd Dec 22 '11 at 20:32
Well, my hint shows that $|Aut(G)|\leq |Aut(H)|$. Can you now show yourself that $|Aut(H)| \leq |Aut(G)|$ (the method is analogous)? Also, i don't directly prove that $Aut(G) \cong Aut(H)$. This however also implies that they have the same size though (see en.wikipedia.org/wiki/Cardinality). – sxd Dec 22 '11 at 21:38