Let $G$ be a group acting on a locally finite tree $T$. Then the boundary $\partial T$ is a Cantor set on which $G$ acts by homeomorphisms (indeed by quasi-isometries under a suitable metric). However, even if $G$ is the full automorphism group of $T$, we can't get the full quasi-isometry group of $\partial T$, as the tree structure puts further restrictions on the action.


What ways are known of identifying those group actions by homeomorphisms on the Cantor set which arise as actions on the boundary of a locally finite tree? Is there a nice algebraic description of $\mathrm{Aut}(T)$ as a subgroup of $\mathrm{Homeo}(\partial T)$?


Edit: Here is an example of a sufficient criterion to show what I mean. Suppose that $G$ acts on the Cantor set $X$ by homeomorphisms, with finite orbits on the clopen subsets. Then one can produce a tree $T$ such that $G$ acts on $T$ with a global fixed point and $\partial T$ is $G$-homeomorphic to $X$. So an action with finite orbits on the clopen subsets is automatically 'arboreal' (we do not care about which tree appears, just that it exists).



Recognising group actions on trees from the boundary
Besides our paper that Alessandro mentions, Sageev, Whyte and I also have a paper "Maximally Symmetric Trees" which comes pretty close to doing what you ask. For certain trees it gives nice descriptions of $Aut(T)$ in terms of the representation into the quasi-isometry group $QI(T)$ (which, by the paper "Un groupe hyperbolique est determine par son bord" of Paulin, is characterized as the quasiconformal subgroup of $Homeo(\partial T)$ in the appropriate sense). 


Although our paper applies only indirectly to all the trees $T$ relevant to your question, it applies very directly to all those trees which are "maximally symmetric", meaning that any continuous, proper, cocompact monomorphism from $Aut(T)$ into $Aut(T')$ for some other tree $T'$ is an isomorphism. Furthermore, the paper describes an explicit way to "expand" any tree $T$ to a maximally symmetric tree $T'$ (equivalently, to embed $Aut(T)$ continuously, properly, and cocompactly in $Aut(T')$), and it gives an explicit enumeration of maximally symmetric trees. The main theorem says that $T$ is maximally symmetric if and only if $Aut(T)$ is a maximal uniform cobounded subgroup of $QI(T)$. 


In all of this, I should clarify by saying that the trees in question are always finite valence trees on which the automorphism group acts cocompactly.


ADDED LATER: I'll admit that this characterization of $Aut(T)$ is more geometric than algebraic, but it at least has well-known analogues in quasi-isometric geometry. For instance it is the analogue of the fact that $Isom(H^3)$, the isometry group of hyperbolic 3-space, is a maximal uniform cobounded subgroup of $QI(H^3)$ (the latter being identified with $Conf(S^2)$, the group of conformal homeomorphisms of the 2-sphere at infinity). The difference is that up to conjugacy there is only one maximal uniform cobounded subgroup of $QI(H^3)$, whereas maximally symmetric trees give countably many different conjugacy classes of maximal uniform cobounded subgroups of $QI(T)$.


