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 $G$).

  • $\begingroup$ 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. $\endgroup$ Dec 22, 2011 at 23:29

1 Answer 1


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?

  • $\begingroup$ if $h \in Aut(G)$ and $\varphi^{-1} \in Aut(H)$implies $\varphi\circ h \circ \varphi^{-1} \in Aut(H)$why? $\endgroup$ Dec 22, 2011 at 20:22
  • $\begingroup$ $\varphi^{-1}$ 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 $H$? $\endgroup$
    – sxd
    Dec 22, 2011 at 20:32
  • $\begingroup$ yes,thanks Dimitri your proof is really impressing $\endgroup$ Dec 22, 2011 at 20:50
  • $\begingroup$ Thanks, no problem, consider accepting it if you like it that much:). $\endgroup$
    – sxd
    Dec 22, 2011 at 21:04
  • $\begingroup$ 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). $\endgroup$
    – sxd
    Dec 22, 2011 at 21:38

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