Factoring isomorphisms using automorphisms 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.
 A: 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.
