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Let $G$ be a finite group . Let $\Gamma$ be the graph $\Gamma=\operatorname{Cay}(G,H)$. Does $\operatorname{Aut}(\Gamma)$ contain an element of order $n=|G|$?

By $\operatorname{Cay}(G,H)$ I mean the Cayley graph of a group $G$ and connecting set $H$ and $\operatorname{Aut}(\Gamma)$ is the automorphism group of the graph $\Gamma$. We mean by "order of $a$" the least $n$ such that $a^n=1_{\operatorname{Aut}(\Gamma)}$.

My question is: when does $\operatorname{Aut}(\Gamma)$ contain a cyclic subgroup of order $n=|G|=|V(\Gamma)|$, when $G$ is nonabelian?

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    $\begingroup$ Consider some graphs that you know have an automorphism like this (complete graphs and cycle graphs come to mind). Can they be a Cayley graph? $\endgroup$ – Morgan Rodgers Jul 5 '17 at 18:17
  • $\begingroup$ Hi Yes complete graphs and cucle craphs are Cayley graphs. $\endgroup$ – Ali Sltan Jul 5 '17 at 21:47
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In general, no. Most groups $G$ admit a connection set $H$ such that $\mathrm{Aut}(\mathrm{Cay}(G,H))\cong G$. (Such a graph is called a graphical regular representation.) These will be counterexamples, unless $G$ is cyclic.

There are other examples, take for example the cube on $8$ vertices. It's Cayley graph (on a few groups), but its automorphism does not contain an element of order $8$.

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