When are toral orbits in buildings the difference of fixed-sets? 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.
 A: (Editted: a "weaker" example about $GL_3$ at the end)
If I didn't make a mistake in my computation, then the second question doesn't hold for $G=Sp_4$, as it doesn't hold for any $\mathbb{Q}_p$. Allow me to use $F$ as my p-adic field and $k$ its residue field. Let $V/_F$ be spanned by $e_2,e_1,f_1,f_2$ with the symplectic form $(e_i,f_j)=\delta_{ij}$, $(e_i,e_j)=(f_i,f_j)=0$, so that we identify $G(F)=Sp(V)$.
The hyperspecial vertices on the building correspond to self-dual lattices of $V$. Let $T$ be the diagonal torus which acts on $V$ by $(c_1,c_2):e_i\mapsto c_ie_i, f_i\mapsto c_i^{-1}f_i$. Let $\pi\in\mathcal{O}_F$ be a fixed uniformizer. Consider the lattices
$$\Lambda_{a,b}=\mathcal{O}_F\langle e_2+\pi^{-1} e_1+\pi^{-2}af_1+\pi^{-3}bf_2, e_1+\pi^{-1}f_1+\pi^{-2}(a-1)f_2,f_1-\pi^{-1}f_2,f_2\rangle,$$
where we let $a,b$ runs over a fixed set of representatives of $k$ in $\mathcal{O}_F$. One checks that all $\Lambda_{a,b}$ are in different $T(F)$-orbits by essentially showing that they are in different $T(\mathcal{O}_F)$-orbits.
Next one compute the stabilizer of $\Lambda_{a,b}$. What's necessary then is to check the following:
Claim. (1) $\text{Stab}_{T(\mathcal{O}_F)}(\Lambda_{a,b})\supset T_3=T(1+\pi^3\mathcal{O}_F)$.
(2) If $ab+b-a+1\not\in\pi\mathcal{O}_F$, then $\text{Stab}_{T(\mathcal{O}_F)}(\Lambda_{a,b})\subset T_2=T(1+\pi^2\mathcal{O}_F)$.
However, there are only $q+3$ ($q=\#k$) subgroups between $T_2$ and $T_3$. In other words, we have $q^2-q+1$ such lattices, but only $q+3$ possible choices for stabilizers for them; an open compact subgroup of $T(F)$ have to correspond to about $O(q)$ orbits.
Note. I did some brute force (using Iwasawa decomp.) and it seems that such an assertion holds for $GL_3$ in most or all cases of $GL_3$. However it also seems to me that if one fix a (semisimple) split rank, then it will be difficult for similar assertion to holds for larger $\dim(G)$; in other words, I'd say such an assertion probably only holds for $GL_n$ (and not for any other classical groups or non-split groups), if it does.
--
Editted: Here is an example about $GL_3$ where things fail. Let $T$ again be the diagonal torus. Let $a>0$ be any integer. Consider the two lattices in $F^3$
$$\Lambda_a=\mathcal{O}_F\langle e_1+\pi^{-a}e_2+\pi^{-a}e_3,e_2,e_3\rangle$$
$$\Lambda_a'=\mathcal{O}_F\langle e_1+\pi^{-a}e_3,e_2+\pi^{-a}e_3,e_3\rangle$$
These two lattices correspond to two hyperspecial vertices which are not in the same $T(\mathcal{O}_F)$-orbit and thus $T(F)$-orbit. But they have the same stabilizer $Z(F)\cdot T(1+\pi^a\mathcal{O}_F)$.
This is however a weaker example because in this case a stabilizer correspond to $2$ orbits. It's probably the case that for $GL_3$ every possible stabilizer corresponds to at most $2$ orbits, but I have no good intuition about why it should be true. (Note that the number of orbits with distant $r$ from the apartment no longer have a uniform bound independent of $p$.)
