Given the data of a triple $(G,h,k)$ where $G$ is a finite group, and $h,k\in G$ of the same order which together generate $G$, I'm interested in understanding the possible pairs $(i,\alpha)$, where $i : G\hookrightarrow \tilde{G}$ is an injection, and $\alpha\in\tilde{G}$ satisfying


1. $\tilde{G}$ is a finite group.
2. There is an $\alpha\in \tilde{G}$ such that $\tilde{G} = \langle G,\alpha\rangle$, and
3. $\alpha h\alpha^{-1} = k$.


There is a natural way to do this, which is [to embed $G$ inside the symmetric group $S\_G$ on $G$ via the left regular representation](https://groupprops.subwiki.org/wiki/Isomorphic_iff_potentially_conjugate_in_finite). I'll translate their proof into my situation: Let $H := \langle h\rangle$, $K := \langle k\rangle$. Now let $\sigma$ denote a pair of transversals $x\_1,\ldots, x\_n$ for $G/H$, and $y\_1,\ldots,y\_n$ for $G/K$, such that $y\_1 = x\_1 = 1$. Now define
$$\alpha\_\sigma : G\rightarrow G \qquad\text{by}\qquad \alpha\_\sigma(h^jx\_i) = k^j y\_i$$
In particular, we have $\alpha\_\sigma(h) = k$.


Then, if $\ell\_t$ for $t\in G$ denotes the permutation given by left-multiplication by $t$, we wish to check that $\ell\_k = \alpha\_\sigma\circ\ell\_h\circ\alpha\_\sigma^{-1}$. To check this, for any $g\in G$, write $g = k^j y\_i$, then note:
$$\alpha(h^jx\_i) = k^jy\_i = g\quad\text{so}\quad h^jx\_i = \alpha^{-1}(k^jy\_i) = \alpha^{-1}(g)$$
Thus,
$$(\alpha\_\sigma\circ\ell\_h\circ\alpha\_\sigma^{-1})(g) = \alpha\_\sigma(h\cdot h^jx\_i)= k^{j+1}y\_i = kg = \ell\_k(g)$$


In particular, letting $L\_G\subset S\_G$ be the image of the left regular representation of $G$, then for any choice $\sigma$ as above, letting $G\_\sigma := \langle L\_G,\alpha\_\sigma\rangle\subset S\_G$, the pair $(G\subset G\_\sigma,\alpha\_\sigma)$ satisfies our desired properties.


Unfortunately, the link above doesn't provide any references for this proof.


**Some questions:**
What is the relation between such $G\_\sigma$ and the infinite HNN extension $\langle G,\alpha | \alpha h\alpha^{-1} = k\rangle$?


Does this construction yield *all* pairs $(i,\alpha)$ satisfying (1),(2),(3)? (E.g., is every finite quotient of $\langle G,\alpha | \alpha h\alpha^{-1} = k\rangle$ isomorphic to some $G\_\sigma$? If so, given a pair $(i,\alpha)$, how does one recover the transversals $\sigma$ and the permutation $\alpha\_\sigma\in S\_G$? If not, is there a more general technique for constructing the ones we're missing?


Does this construction satisfy some kind of universal property?


Does there exist a group $G$ for which any pair of generators of $G$ are conjugate (ie, the two generators are conjugate elements) in $G$?


Are any of these questions answered in the literature? (references would be appreciated).



Classification of finite HNN-extensions of a finite group with respect to an isomorphism between cyclic subgroups
This is probably an extended comment rather than an answer. An HNN extension of a finite group is residually finite, in fact, virtually free. The element $\alpha$ has infinite order in the HNN extension so you can find finite images where it maps to elements of arbitrarily large order. Your construction only produces conjugators of bounded order (in fact you create only finitely many images of the HNN extension). So you are missing many quotients. 


**Added.** Let $H$ be the HNN extension. Then your construction shows that there is a normal subgroup $N$ of finite index such that $G\cap N=\{1\}$ (take the quotient map $H\to G\_{\sigma}$). It follows that $N$ contains no conjugate of an element of $G$. Therefore, the action of $N$ on the Bass-Serre tree of $H$ is free and so $N$ is a free group. Therefore, $H$ has a free subgroup of finite index and hence $H$ is residually finite. Therefore, any finite set of elements of $H$ can be separated by a finite quotient of $H$. In particular, $G$ and any power of $\alpha$ can be separated in a finite image of $H$.


