When are two HNN extensions isomorphic? Let $G$ be a finite group. Consider triples of the form $(G,g,h)$, where $g,h$ are generators of $G$. Let
$$H(G,g,h):= \langle G,t | tgt^{-1} = h\rangle$$
be the HNN extension corresponding to the triple $(G,g,h)$
Suppose we have two triples $(G,g,h)$ and $(G',g',h')$ as above such that $$H(G,g,h)\cong H(G',g',h')$$
What does this tell us about $(G,g,h)$ and $(G',g',h')$?
For example, is it possible that $G\not\cong G'$? If they must be isomorphic, what does this tell us about $g,h$ and $g',h'$?
For example, is it possible that $H(G,g,h)\cong H(G',g',h')$ if and only if there is an isomorphism $\varphi : G\rightarrow G'$ and $z\in G'$ such that $g' = z\varphi(g)z^{-1}$ and $h' = z\varphi(h)z^{-1}$?
 A: First of all, you should read Serre's book "Trees" (unless you already have read it). From this book you learn that amalgams of groups (and HNN extensions) correspond to group actions on simplicial trees (without inversions). For instance, if you have an HNN extension $G$ of $H$ with the stable letter $t$, then for the corresponding $G$-action on a tree $T$, the subgroup $H$ fixes a vertex and all vertex stabilizers will be conjugate to $H$. The stable letter $t$ acts as a translation on $T$, so no fixed vertices. 
A group $H$ is said to have the Property FA if every such action on a tree has a fixed vertex. All finite groups have this property. 
Suppose now that you have two HNN decompositions $HNN(H_1, t_1)$, $HNN(H_2, t_2)$ of a group $G$ such that both $H_1, H_2$ are finite (this will make our life easier). Let $T_1, T_2$ denote the corresponding $G$-trees. Then $H_1$ fixes a vertex in both $T_1$ and in $T_2$. Hence, we obtain that $H_1$ is conjugate to a subgroup of $H_2$ and vice versa. Since $H_1, H_2$ are both finite, it follows that they are conjugate in $G$. In particular, $H_1\cong H_2$. 
One can prove more by looking at the edge-stabilizers. If the HNN extensions $HNN(H_1, t_1)$, $HNN(H_2, t_2)$ are given by $t_1A_1t_1^{-1}=A_1'$, $t_1A_1t_1^{-1}=A_1'$, then the subgroups $A_1, A_2$,  represent the conjugacy classes of the edge-stabilizers in $T_1, T_2$. Hence, up to conjugation (since the vertex stabilizer acts transitively on the incident edges in both $T_1, T_2$), we can conjugate $H_1, H_2$ so that (after conjugation) $H_1=H_2$ and $A_1=A_2$. 
