Let $D$ be a bounded hermitian symmetric domain with automorphism group $G(\mathbb R)$. In the example I have in mind, $D$ is Siegel upper half-space of degree $g$ and $G(\mathbb R) = \mathrm{Sp}(2g,\mathbb R)$.

Let $O$ be an order in a totally real number field. Example: $O=\mathbb{Z}[\sqrt{d}]$ with $d\in \mathbb Z$ positive.

Let $\Gamma$ be a torsionfree finite index normal subgroup of $G(O)$. Fix an embedding $G(O)\subset G(\mathbb R)$. Note that $\Gamma$ acts on $D$.

Are all isotropy groups of $\Gamma$ mutually isomorphic?

The answer is positive if $O = \mathbb Z$, as then the isotropy groups of the action are compact and discrete (hence finite), and thus trivial.

In general, I suspect that the isotropy groups will be free $\mathbb Z$-modules of finite rank. I'm asking whether the rank can be vary with the points of $D$.