This is not an answer but rather a personal thought related to Matt Zaremsky's answer.

Thompson's groups $F$, $T$ and $V$ can be interpreted as groups of *partial isometries* of the $3$-regular tree $A_3$. A partial isometry is the data of two finite subtrees $R,S \subset A_3$ and an isometry $f : A_3 \backslash R \to A_3 \backslash S$. Two partial isometries are considered as equal if they are well-defined and agree on the complement of a finite subtree. Similarly, the Higman-Thompson groups $F_{n,r}$, $T_{n,r}$, $V_{n,r}$ are subgroups in $\mathrm{PIsom}(A_{n+1,r})$ where $A_{n+1,r}$ is the forest having $r$ trees which are all $(n+1)$-regular.

Naturally, a partial isometry of $A_3$ extends to a *quasi-automorphism* (aka *almost isometry*) of $A_3$, i.e. a bijection of the vertices of $A_3$ that preserves adjacency and non-adjacency for all but finitely many pairs of vertices. This leads to extensions $QF$, $QT$, $QV$ of $F$, $T$, $V$, which can be thought of as *symmetrisations* of Thompson's groups because of the exact sequences
$$1 \to S_\infty \to QF \to F \to 1$$
and similarly for $T$ and $V$, where $S_\infty$ denotes the group of finitely supported permutations of a countably infinite set.

**Generalisation.** Now, consider an arbitrary tree $A$ (but infinite, otherwise there is nothing interesting to say). You still have an exact sequence
$$1 \to S_\infty \to \mathrm{QAut}(A) \to \mathrm{PIsom}(A),$$
but the third arrow may not be surjective in general (see Houghton's groups below). So $\mathrm{QAut}(A)$ can be thought of as a symmetrisation of its image in $\mathrm{PIsom}(A)$.

**Houghton's groups.** As a particular case, instead of a regular tree, consider the graph $R_n$ which is a union of $n$ infinite rays with a common origin. Then the group of partial isometries $\mathrm{PIsom}(R_n)$ is isomorphic to $\mathbb{Z}^n \rtimes S_n$ and the group of quasi-automorphisms $\mathrm{QAut}(R_n)$ is isomorphic to $H_n \rtimes S_n$ where $H_n$ denotes the $n$th Houghton group. From the exact sequence above, $H_n$ can be thought of as a symmetrisation of $\mathbb{Z}^{n-1}$.

Of course, $H_n$ is quite different from Thompson's groups, but I think that the point of view described here explains the similarity in the techniques which can be used to study both Houghton's and Thompson's groups.

**Lamplighter groups.** Another example I find interesting is when $A$ is a bi-infinite horizontal line $\mathbb{R}$ with an infinite vertical descending ray attached at each integer. Then $\mathrm{PIsom}(A)$ is isomorphic to $\left( \bigoplus\limits_{n \in \mathbb{N}} \mathbb{Z} \right) \rtimes S_\infty$. In the same way that $T$ lies in $\mathrm{PIsom}(A_3)$, the lamplighter group $\mathbb{Z} \wr \mathbb{Z}$ lies in $\mathrm{PIsom}(A)$. By looking at the pre-image under $\mathrm{QAut}(A) \to \mathrm{PIsom}(A)$ of a subgroup isomorphic to $\mathbb{Z} \wr \mathbb{Z}$, it is possible to introduce a symmetrisation of $\mathbb{Z}\wr \mathbb{Z}$. Up to my knowledge, this example does not appear in the literature, but this is probably because such a group is not finitely presented.

A symmetrisation of $\mathbb{Z} \wr \mathbb{Z}$ should be compared to $H_2$, and a problem I am interested about is: similarly, what would be a good analogue of $H_n$? It should be a (free abelian)-by-$\mathbb{Z}^{n-1}$ group which is of type $F_{n-1}$ but not of type $F_n$.

Anyway, if $H_n$ is considered as an amenable Thompson-like group, so should be the lampligher group $\mathbb{Z} \wr \mathbb{Z}$.

non-amenable. $\endgroup$ – YCor Jul 8 '18 at 13:47