Consider a permutation group $G$ acting on an infinite set $X$. Assume $G$ has *finitely many* orbits, and every point stabiliser $G\_x$ has *finite* orbits.


Can we always find a permutation $\tau\in\operatorname{Sym}(X)$ (not necessarily of finite order or finite support) such that $H=\langle G,\tau\rangle$ is transitive, while every point stabiliser $H\_x$ still has finite orbits?


As noted in [this previous question](https://mathoverflow.net/questions/311474) some obvious choice of $\tau$ will not work in general.



Enlarging a subdegree-finite "almost transitive" permutation group to a transitive one? (follow-up)
It seems that a counterexample looks as follows (it is somewhat siimilar with the counterexample to your previous question). Let $k$ be a large integer. Take an infinite tree $T=(V,E)$, where all degrees equal $k+1$. Let $G$ be its group of automorphisms. $G$ acts on $V\cup E$ with two obvious orbits, and all orbits of the stabilizers are finite.


I could find only quite technical proof that this example works; prehaps, there are easier ones.


Define the metric $d(\cdot,\cdot)$ on $V\cup E$ identifying each edge with its midpoint (the length of every edge is $1$). E.g., the distance between a vertex and an incident edge is $1/2$.


Assume that we took a transitive group $H\geqslant G$ with finite stabilizers' orbits. Let $G\_v$ and $G\_e$ be stabilizers of a vertex $v$ and an edge $e$ in $G$, and $H\_v$ and $H\_e$ be those in $H$. Then $H\_e$ and $H\_v$ are conjugates, so there is a cardinality-preserving bijection of their sets of orbits. Say that the *radius* of $H\_v$-orbit is the maximal distance from $v$ to its element; the same for $H\_e$-orbits.


$H\_v$-orbits are unions of $G\_v$-orbits which have cardinalities $(k+1),(k+1),k(k+1),k(k+1),\dots,k^n(k+1),k^n(k+1),\dots$. Similarly, $H\_e$-orbits are unions of $G\_e$-orbits whose cardinalities are $2,2k,2k,2k^2,\dots,2k^n,2k^n,\dots$ (notice here that an $H\_v$-, and hence $H\_e$-, orbit cannot have just $2$ elements). So the cardinality of any $H\_v$-orbit $\Omega\_v$ has the form either $k^n+O(k^{n-1})$ or $2k^n+O(k^{n-1})$, and that of the corresponding $H\_e$-orbit $\Omega\_e$ may have the form either $2k^n+O(k^{n-1})$ or $4k^n+O(k^{n-1})$. Hence this (common) cardinality is $2k^n+O(k^{n-1})$. This means that the radius of $\Omega\_v$ is an integer $r$< and that of $\Omega\_e$ is either $r$ or $r+1/2$.


*Case 1.* $\Omega\_v$ and $\Omega\_e$ have the same radius $r$. 
Consider now some $\tau\in H$ mapping $v$ to $e$; every $H\_v$-orbit $\Omega\_v$ is mapped to some $H\_e$-orbit $\Omega\_e$ of the same radius $r$. Notice that a dominating part $\partial\Omega\_v$ of $\Omega\_v$ consists of *far vertices* $v'$ with $d(v,v')=r$ and *far edges* $e'$ with $d(v,e')=r-1/2$; similarly, a dominating part $\partial \Omega\_e$ of $\Omega\_e$ consists of *far edges* $e'$ with $d(e,e')=r$. So most of such permutations $\tau$ map a good proportion of far vertces and edges (for $v$) to far edges (for $e$). See footnote for the explanation of the term `most of'.


Now consider some orbit' radius $r$. Take a vertex $v$ ans some $\tau$ mapping $v\mapsto e$. Let $\Omega\_v$ and $\Omega\_e$ be $H\_v$- and $H\_e$-orbits of radius $r$. Under $\tau$, most of $k^{r-1}(k+1)$ far edges for $v$ map to far edges for $e$. 


Let $e'$ be an edge incident to $v$, and let $\Omega\_{e'}$ he $H\_{e'}$-orbit of radius $r$. Notice that $k^r$ of far edges for $e'$ are also far edges for $v$. So, most of them also map to far edges/vertices for $\tau(e')$, so thare are almost $k^r$ common far edges for $e$ and $\tau(e')$. This may happen only if $\tau(e')$ is a vertex incident to $e$. But $e'$ can be chosen in $k+1$ ways, while $e$ has only two endpoints. A contradiction.


*Case 2.* Assume that the radius of $\Omega\_e$ is $r+1/2$. By symmetry, we may assume that $d(v,e)=r+1/2$, so that $v\in\Omega\_e$. Due to the cardinalities, $\Omega\_e$ should contain also either all vertices at distance $r-1/2$ from $e$, or all edges $e'$ at distance $r-1$ from $e$. In the former case, we get that an edge at distance $r-1/2$ and an edge at distance $r+1/2$ are equivalent modulo $H\_v$, so $e\in\Omega\_v$, which is impossible. In the latter case, we have $\sigma\in H$ with $\sigma(e')=v$, $\sigma(e)=e$. But, since $e$ lies in $\Omega\_{e'}$ corresponding to $\Omega\_e$, we get $e\in\Omega\_v$ again.


*Footnote.* "Most of permutations $\tau$" is regarded in the following sense. If we consider all permutations $\tau$ mapping some $x\mapsto y$, there are only finitely many $t$ into which may a fixed $z$ map. If we are interested in images of a finite set of such $z$'s, there are finitely many tuples of images, and they are `equally distributed' (since all $\tau$ realizing one tuple form a coset of their joint stabilizer). Now we may speak on the probability in this sense.


