Is there a way to construct a "product" of graphs $G\rtimes H$ such that $Aut(G \rtimes H ) \simeq Aut(G) \rtimes Aut(H) $? A related topic is "Semidirect product" of graphs? but not quite the same.

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    $\begingroup$ What action of the group $\text{Aut}(H)$ on the group $\text{Aut}(G)$ did you have in mind for defining their semidirect product? $\endgroup$
    – Lee Mosher
    Commented May 16, 2015 at 13:42
  • $\begingroup$ Any abstract action, the semi direct product of the graphs will depend on this action. $\endgroup$ Commented May 17, 2015 at 8:36
  • $\begingroup$ For example if I want to find a graph with automorphism group isomorphic to $\mathbb{Z}_3 \rtimes \mathbb{Z}_9$ I don't have to follow Frucht's construction instead I can use this method. And contraction of graphs with automorphisms $\mathbb{Z}_3$ and $\mathbb{Z}_9$ is easier. $\endgroup$ Commented May 18, 2015 at 3:05
  • $\begingroup$ Do you mean the group product? What type of product do you mean? $\endgroup$ Commented May 18, 2015 at 13:17
  • $\begingroup$ Given an action of $ Aut(H )$ on $ Aut(G)$ I want a product to be defined on the two graphs $G$ and $H$ such that the automorphism group of this product is the semidirect product of the automorhpism groups of $G$ and $H$ under the action. $\endgroup$ Commented May 18, 2015 at 13:34


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