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

$\tilde{G}$ is a finite group.

There is an $\alpha\in \tilde{G}$ such that $\tilde{G} = \langle G,\alpha\rangle$, and

$\alpha h\alpha^{-1} = k$.

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. 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).