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Let $H$ be any group and $K$ an abelian group. (I'm interested in $K={\mathbb Z}$.) Homomorphisms $H\to Aut(K)$ define semi-direct products $K\rtimes H$.

There is an action of $Aut(H)\times Aut(K)$ on $Hom(H,Aut(K))$, where $Aut(K)$ acts on itself by conjugation. Homomorphisms in the same $Aut(H)\times Aut(K)$-orbit yield isomorphic semi-direct products. (Isomorphic as a group, of course not as an extension.)

I'm interested in a converse. In the theory of finite groups it is known that under the assumption $gcd(\mid H\mid,\mid K\mid)=1$ two semi-direct products are isomorphic only if the corresponding homomorphisms $H\to Aut(K)$ belong to the same $Aut(H)\times Aut(K)$-orbit. (Unfortunately I don't have a reference for this fact.)

My question is for $K={\mathbb Z}$ and infinite groups $H$. (Say free groups or surface groups.) Can there be isomorphic semi-direct products in different $Aut(H)\times Aut(K)$-orbits?

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    $\begingroup$ For this to happen, the semidirect product would have to have a second normal subgroup isomorphic to ${\mathbb Z}$, so $H$ would have to have a normal cyclic subgroup, which is not true for nonabelian free groups and surface groups. I am doubtful whether it is possible for any group $H$, but that would need more thought. $\endgroup$ – Derek Holt Jul 27 '14 at 11:00
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    $\begingroup$ There is a paper of Arzhantseva, Lafont and Minasyan, Isomorphism versus commensurability for a class of finitely presented groups, where they take $H\cong\mathbb{Z}$ and $K$ not mapping onto $\mathbb{Z}$ and they give necessary and sufficient conditions for two such semidirect products to be isomorphic. This doesn't answer your question, but you might find it interesting all the same... $\endgroup$ – user1729 Jul 29 '14 at 10:40

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