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