Let $L$ be a $p$-adic field, let $G$ be a reductive group over $L$ (I'm even okay assuming semisimplicity for now). Let $T$ be a maximal torus of $G$. Let $B$ be the building for $G(L)$. (**Edit 1:** "split" has been removed from "maximal torus"; the building of $U_3(L/F)$, where $L/F$ is a ramified quadratic extension of $p$-adic fields, provides a counterexample to the question if we require $T$ to be split).

I'm interested in the relationship between, on the one hand: $T(L)$-orbits in $B$ and, on the other hand: fixed sets $B^{T_0}$ where $T_0 \subseteq T(L)$ is an open-compact subgroup.

Clearly, since $T_0$ is commutative, any set $B^{T_0}$ is closed under multiplication by $T$ and is therefore a union of $T(L)$-orbits. It is, of course, too much to hope that every $T(L)$-orbit is of the form $B^{T_0}$ for some fixed compact-open subgroup $T_0$ of $T(L)$, for two reasons: the first is that all elements of a $T(L)$-orbit are of the same type, which is not true of sets of the form $B^{T_0}$. The second is that for any open-compact subgroup $T_0$, $B^{T_0}$ contains the apartment of $T(L)$ (and possibly contains more).

My question is as follows:

Let $X$ be a $T(L)$-orbit in $B$, and let $Typ(x) = t$ for every $x\in X$. When are there compact-open subgroups $T_0 \subseteq T_0'$ such that $X = (B^{T_0} - B^{T_0'})\cap Typ^{-1}(t)$?

and, more, specifically:

For a reductive (semisimple) group over a global field $F$, are there only finitely many primes at which the above fails?

Let me give a little bit of clarity by way of example. Let's take $G = SL_2$, let $T$ be the diagonal torus, and let $A_0$ be the standard apartment. Then one can check by direct computation that two vertices $v,\, v'$ are in the same $T$-orbit if they are of the same type and if $d(v, \,A_0) = d(v', \,A_0)$ (i.e. they are the same distance from the standard apartment).

On the other hand if define the open-compact subgroups $$T_{r} := \bigg\{\begin{pmatrix} t & 0 \\ 0 & t^{-1}\end{pmatrix}: t \in 1 + \varpi^r \mathcal{O}_L\bigg\}.$$ Then if the residue characteristic of $L$ is at least $3$, the fixed set $B^{T_r}$ consists of all vertices that are at most distance $r$ from the standard apartment $A_0$. As such, in the case where the residue characteristic is $> 2$, we can answer in the affirmative; if an orbit $X$ consists of vertices that are distance $r$ from $A_0$, then we can write $X = B^{T_r} - B^{T_{r-1}}$.

On the other hand if the residue characteristic of $L$ is $2$, and $T_0$ is the maximal compact subgroup of $T_0$, then by the notation above $T_0 = T_1$ so $T_0$ fixes the standard apartment and all vertices of distance $1$. Therefore, we cannot write the standard apartment as a difference of fixed sets.

This question is extremely open-ended, and I'm really interested in anything that can be said in this area. Even some illumination of examples where $G$ is split-rank $1$ (so $B$ is a tree) beyond $SL_2$ would be of great interest to me.