TITLE: Infinite groups with oligomorphic conjugation action
QUESTION [16 upvotes]: The action of a group $G$ on a set $X$ is called oligomorphic if the diagonal action on $X^n$ has finitely many orbits for each $n$.

Question: Is there an infinite (maybe even finitely generated) group $G$ such that the conjugation action of $G$ on itself is oligomorphic?

Edit: In view of the comments, I would already be happy to see an example of an infinite group $G$ whose conjugation action on $G^2$ or $G^3$ has finitely many orbits.

REPLY [8 votes]: Another argument to show that a conjugation-oligomorphic group G is finite is as follows. Let $(s_i,t_i)$ be a set of representatives of the $G$-action on $G\times G$ and let $X$ be the subgroup generated by these representatives. Note that whenever we consider a subgroup generated by $X$ and two more elements $a,b$, $Y=\langle X,a,b\rangle$, then $Y$ is a subgroup of $\langle X,t\rangle$ where $t$ is such that it conjugates $(a,b)$ to some $(a_i,b_i)$.
Continuing this process, i.e. passing from $\langle X,a,b\rangle$ to $ \langle X,t\rangle$ to $\langle X,t,u\rangle$ to $\langle X, v\rangle$ (where $v$ is such that it conjugates $(t,u)$ into $X$)...  we get an increasing chain of $(m+1)$-generated subgroups. As $G$ is uniformly locally finite (see the comments or (3) in YCor's answer), $G$ must be finite.