Can Thompson's group F be realized as a semigroup of continuous transformations of a tree? I am not very familiar with F, but I know that it can be realized as a group of homeomorphisms of the boundary of the binary tree.  I also know that F cannot be realized as a group of graph automorphisms of any regular rooted tree because F is not residually finite.  However, if we topologize our trees with the path metric, can F be realized as a group of continuous prefix-preserving transformations of a regular rooted tree (where the transformations need not be injective or surjective)?  If you know the answer, could you provide a reference?  Thanks!
 A: A semigroup of level preserving transformations of a rooted tree is still residually finite.  The levels are still finite and so the actions on the levels separate points into finite semigroups.  Thus no such representation exists.
A: There is a way to get $F$ to act on the infinite binary tree bijectively, but I doubt it satisfies most of your other requirements.  It basically does something sensible with the "missing finite subtree."  I have only partly checked this out (meaning it seems to check for one of the two relations needed).
We let $T$ be the set of finite words (including the empty word) on the alphabet $\{0,1\}$.  This is a binary tree by letting the left child of $u\in T$ be $u0$ and the right child of $u$ be $u1$.  We define two permutations of $T$.
The permutation $x_0$ is determined by the following rules:
$\emptyset\rightarrow 1$; 
$0\rightarrow \emptyset$;
$00u\rightarrow 0u$;
$01u\rightarrow 10u$;
$1u\rightarrow 11u$.
The permutation $x_1$ is determined by the following rules:
$\emptyset\rightarrow \emptyset$;
$0u\rightarrow 0u$;
$1\rightarrow 11$;
$10\rightarrow 1$;
$100u\rightarrow 10u$;
$101u\rightarrow 110u$;
$11u\rightarrow 111u$.
These are the usual rules for the action of $x_0$ and $x_1$ in $F$ on infinite words in $\{0,1\}$ restricted to finite words and extended to the few cases that the rules usually omit.
As I said, it checks for the relation:
$(x_1)^{x_0x_0} = (x_1)^{x_0x_1}$.
Here $a^b$ means $b^{-1}ab$ and the actions are to be composed from left to right (they are right actions).
The other relation that defines $F$ with the one above is
$(x_1)^{x_0x_0x_0} = (x_1)^{x_0x_0x_1}$.
If the second fails while the first succeeds, I will be stunned.
Assuming that the second relation checks out (not too hard, I am just too lazy), then these two permutations of $T$ generate a copy of $F$.  On "most" of $T$, the action agrees with the usual action.  How well this cooperates with what you want is for you to decide.
The definitions can be tinkered with a bit.  I doubt that the relations can survive a lot of tinkering though.
A: There is a so-called group of hierarchomorphisms $\mathsf{Hier}(T)$ of a homogeneous tree $T$ introduced by Neretin. It consists of homeomorphisms of the boundary $\partial T$ which can be extended to $T$, except for a finite subtree, and is, in a sense, similar to the group of diffeomorphisms of the circle. Thompson's group $F$ can be realized as a subgroup of $\mathsf{Hier}(T)$.  
