Let $G$ be a reductive $p$-adic group and let $\mathcal B$ denote its Bruhat-Tits building. For $x \in \mathcal B$, denote by $\overline x$ the (closure of the) minimal facet containing $x$. We associate to $x$ the open compact subgroup $G_x$ consisting of all elements $g\in G$ such that $g\cdot \overline x = \overline x$.
Assume that for two points $x,y \in \mathcal B$, we have $G_x = G_y$. Is it true that $\overline x = \overline y$ ?
I think it is known that $G_x$ acts transitively on the set of all appartments in $\mathcal B$ which contain the facet $\overline x$. Thus, our assumption implies that $\overline x$ and $\overline y$ live in precisely the same appartments. However, I don't see how I could go further to prove the equality (assuming that it is true).