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The underlying question is: how closely related are isomorphisms and automorphisms? More precisely, if $G_1$ and $G_2$ are given in terms of generating sets, when can we factor an isomorphism as an automorphism and an "obvious" map.

Let $H=\langle S\rangle$ be a fixed group, and let $G_1=H\ast_{A_1^{t_1}=B_1}$ and $G_2=H\ast_{A_2^{t_2}=B_2}$. Note that $G_1=\langle S, t_1\rangle$ and $G_2=\langle S, t_2\rangle$. If $\alpha:G_1\rightarrow G_2$ is an isomorphism then does there exists an automorphism $\beta\in\operatorname{Aut}(G_2)$ such that $\alpha$ factors as $\beta\circ \operatorname{id}$, where $\operatorname{id}: G_1\rightarrow G_2$ is the map with $\operatorname{id}(s)=s$ for all $s\in S$ and $\operatorname{id}(t_1)=t_2$.

In my precise setting, $H$ has Serre's property FA. Note that there are some subtleties here with respect to generating sets. For example, if $H=\mathbb{Z}$ then $BS(2, 3)=\langle a, t; t^{-1}a^2t=a^3\rangle$ is generated both by the pair $(a, t)$ and by $(a^2, t)$, but there is no automorphism of $BS(2, 3)$ which takes one pair to the other (this is the classical Baumslag-Solitar map, with non-trivial kernel). This observation is a subtlety rather than an outright obstruction though: $(a^2, t)$ is not given as an HNN-extension unlike in the question.

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    $\begingroup$ If $\alpha: G_1 \rightarrow G_2$ is an isomorphism, then every isomorphism from $G_1$ to $G_2$ is of the form $\beta \circ \alpha$ where $\beta \in \mathrm{Aut}(G_2)$. So the only question is whether mapping one generating set to the other extends to an isomorphism, which sounds like a hard problem in general (but maybe feasible for some special kinds of presentation). $\endgroup$ – Colin Reid Aug 23 '17 at 12:17
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    $\begingroup$ @Colin In my specific case I want to show non-isomorphism. I know that mapping one generating set to the other is not an isomorphism, and I want to know if I can stop there. My question is whether is it feasible for HNN-extensions; whether they are special enough. $\endgroup$ – user113727 Aug 25 '17 at 6:47
  • $\begingroup$ It took me a while to notice your restriction that $H$ has Serre's property FA. Before I noticed that, I was ready to write out the following: For the case where $H$ is the fundamental group of a closed, oriented surface of genus $g \ge 1$, there are many counterexamples arising from the Thurston norm (see this very closely related answer mathoverflow.net/questions/104873/…). But I suppose you knew that. $\endgroup$ – Lee Mosher Sep 25 '17 at 21:09
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I think you can mostly get what you want if in addition, you assume $\mathrm{Out}(H)$ is trivial and one of the HNN extensions is not ascending (let's say $G_1$).

Suppose $\alpha: G_1 \rightarrow G_2$ is an isomorphism. Then via $\alpha$, $G_1$ acts on the Bass–Serre tree for $G_2$ and vice versa. By Serre's property (FA), $\alpha(H) \le gHg^{-1}$ for some $g \in G_2$ and $\alpha^{-1}(gHg^{-1}) \le hHh^{-1}$ for some $h \in G_1$. So we end up with $H \le \alpha^{-1}(gHg^{-1}) \le hHh^{-1}$. Since it's a nonascending HNN extension, $H$ is not conjugate in $G_1$ to any proper overgroup of itself, so $H = \alpha^{-1}(gHg^{-1}) = hHh^{-1}$; after conjugating in $G_2$ we can then assume $\alpha(H) = H$. Since $\mathrm{Out}(H)$ is trivial, actually $\alpha$ is then inner on $H$, and by conjugating again in $G_2$, we get an isomomorphism from $G_1$ to $G_2$ that restricts to the identity on $H$.

On the other hand, if $\mathrm{Out}(H)$ is nontrivial, you could have $G_1 = H \ast_{A^{t_1}_1 = B_1}$, $G_2 = H \ast_{(\beta(A_1))^{t_2} = \beta(B_2)}$, where $t_2 =\beta t_1 \beta^{-1}$ for some noninner automorphism $\beta$ of $H$. In that case $G_1$ and $G_2$ are certainly isomorphic, but probably not in a way that restricts to the identity on $S$.

I'm not sure what happens in the case when $G_1$ and $G_2$ are both ascending.

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  • $\begingroup$ I do not understand why $\alpha(H)=H$ implies that $\alpha$ is inner. For example, in $BS(2, 4)=\langle a, t; t^{-1}a^2t=a^4\rangle$ the automorphism $\psi: a\mapsto a, t\mapsto t^{-1}a^2ta^{-2}t$ acts trivially on the base group $\langle a\rangle$ but is not inner. (I know that this isn't the precise setting you wrote, but I am not convinced that $\alpha$ is inner in your setting, and this example illustrates what I think the issue is.) $\endgroup$ – user113727 Aug 25 '17 at 16:01
  • $\begingroup$ I mean it's inner on $H$ (edited). I don't think you can control where the stable letter will go: given a group $G = H \ast_t$, then typically $G$ will also split as $G = H \ast_s$ for many other elements $s$. $\endgroup$ – Colin Reid Aug 26 '17 at 17:38

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