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.
There is no such group.
Say that $G$ is $n$-CO (for "conjugation-oligomorphic") if $G$ has finitely many orbits on $G^n$ by diagonal conjugation $g\cdot (g_1,\dots,g_n)=(gg_1g^{-1},\dots)$. (Clearly this implies $(n-1)$-CO). Say that $G$ is CO if it is $n$-CO for all $n$. Define ACO in the same way, but for the action of the automorphism group $\mathrm{Aut}(G)$. So $n$-CO implies $n$-ACO and CO implies ACO. (ACO is better known as "$\aleph_0$-categorical" or "$\omega$-categorical".)
(1) If $G$ is 1-ACO then torsion elements in $G$ have bounded order.
(2) If $G$ is 2-ACO then $G$ is torsion (because if $g$ is non torsion, the pairs $(g,g^n)$, for $n\ge 2$, are in distinct orbits for the action of $\mathrm{Aut}(G)$. In particular, in combination with (1), 2-ACO implies torsion of bounded exponent.
(3) (generalization of 2): if $G$ is $(n+1)$-ACO, then $n$-generated subgroups of $G$ are (uniformly) finite: indeed if $(g_1,\dots,g_n)\in G^n$, then the $(g_1,\dots,g_n,h)$, for $h$ ranging over elements of the subgroup $\langle g_1,\dots,g_n\rangle$ are in distinct orbits of $\mathrm{Aut}(G)$. In particular, if $G$ is ACO then $G$ is locally finite.
(4) If $G$ is $n$-CO (or CO), then so is every finite index normal subgroup $H$ of $G$ (because transitive $G$-actions have finitely many $H$-orbits, and this applies to the $G$-action on $H^n$).
(5) If $G$ is 1-CO then $G$ has finitely many normal subgroups. In particular, it has a unique minimal subgroup of finite index $G^\sharp$. If $G$ is $n$-CO (or CO) then so is $G^\sharp$ (by (4)).
(6) If $G$ is $n$-CO (or CO) then so is every quotient of $G$. If $G\neq 1$ then (by (5)) it has a maximal normal proper subgroup, and hence a simple quotient (which is thus $n$-CO or CO as well).
(7) By (5) and (6), if $G$ is infinite and $n$-CO (or CO), then $G$ has a subquotient $H$ that is an infinite simple group with $n$-CO (or CO).
(8) If $G$ is infinite CO (or $n$-CO) then $G$ has an infinite countable subgroup $H$ that is CO (or $n$-CO).
[Let us check only the case of CO, the others are slightly easier. Start with any infinite countable subgroup $H_0$. Enlarging $H_0$ allows to assume that for every $n$, every $n$-tuple in $G^n$ is conjugate to some finite subset $X_n$ of $n$-tuples contained in $H_0^n$. Next, we use that for every countable subgroup $L$ and $n$, adding countably many elements embeds $L$ in a countable subgroup $M$ such that all $n$-tuples in $L^n$ are conjugate by $M$ to an $n$-tuple in $X_n$. Then we use this for $n=1$ to $H_0$ to get $H_0\subset H_1$, then to $H_1$ and $n=2$, etc: every $n$-tuple in $H_{n-1}$ is conjugate in $H_n$ to an element in $X_n$. In particular, it follows that $H_\infty=\bigcup H_n$ is CO: indeed if $(g_1,\dots,g_n)\in H_\infty^n$, it belongs to $H_{m-1}^n$ for $m$ large enough, say $m\ge n$. Then the $m$-tuple $(g_1,\dots,g_n,1,\dots,1)$ belongs to $H_{m-1}^m$ and hence is conjugate by $H_m$ (and thus by $G$) to an element of $X_m$. Assuming (as we can that $X_n\times\{1\}^{n-m}$ is contained in $X_m$ for all $n\le m$, we deduce that $(g_1,\dots,g_n)$ is conjugate by $H_\infty$ to an element of $X_n$.]
Now I will use some known results:
(9) If $G$ is locally finite, simple and locally solvable, then $G$ is not simple (unless $G$ is cyclic of prime order). See Corollary 3.5 here (original references welcome!).
(10) If $G$ is a countable, characteristically simple group, locally finite group of finite exponent, then either (a) $G$ is a $p$-group for some prime $p$, or (b) $G$ has the form $B(S),B^-(S),P(S)$ for some finite nonabelian simple group $S$. (Reference: ) Here $P(S)$ is the restricted direct product of countably many copies of $S$, $B(S)$ is the group of continuous functions $K\to S$ ($K$ the Cantor set), and $B^-(S)$ is the subgroup of $B(S)$ of those functions vanishing at some $x_0\in K$. Reference: (5) in The algebraic structure of $\aleph_0$-categorical groups. Groups—St. Andrews 1981 (St. Andrews, 1981), pp. 345–358, London Math. Soc. Lecture Note Ser., 71, Cambridge Univ. Press, Cambridge-New York, 1982. (Remark: I found this reference in a Google search on locally finite groups of finite exponent; anyway it's not just a coincidence if it lands on a study of $\aleph_0$-categorical groups, as suggested by Julien Melleray.) [Additional reference (including a proof of Wilson's result): A.B. Apps, On the structure of $\aleph_0$-categorical groups, Journal of Algebra Volume 81, Issue 2, April 1983, Pages 320-339. (Sciencedirect link)]
The groups in (10)(b) are not CO, not even 1-CO (because the support is conjugation invariant).
From (9) and (10) we get
(11) There is no infinite countable simple locally finite group of finite exponent.
Using (8) and then (7), if there is an infinite CO-group, then there is one that is in addition simple, and is locally finite by (3). With (11), this leads to a contradiction.
(Remark: possibly following Wilson's arguments can lead to a more direct proof, without this business to pass to a countable subgroup, when we assume simplicity instead of characteristic simplicity. I haven't yet been through the proof, only relying on a partially scanned Google version.)
This leaves open whether there is an infinite $n$-CO group for any finite value of $n$; even $n=2$ is unclear.
The same argument proves that there is no infinite locally finite 2-CO group, so proving that there is no infinite $n$-CO group is equivalent to proving that every (countable simple, if one wishes) $n$-CO group is locally finite.
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.
