Let $Q$ be the Lorentzian matrix, that is $Q=\mathrm{diag}(I_n,-1)$, where $I_n$ denotes de $n\times n$ identity matrix. Let $\mathcal R$ the set of integer solutions $x\in\mathbb Z^{n+1}$ of $$ Q[x]:=x^tQx=x_1^2+\dots+x_{n}^2-x_{n+1}^2=-1. $$ The set $\Gamma=\mathrm{O}(n,1)\cap \mathrm{M}(n+1,\mathbb Z)$ acts by left multiplication on $\mathcal R$. It's known that there are finitely many $\Gamma$-orbits.

I think (but I'm not completely sure) that $\Gamma\backslash \mathcal R$ has only one class if $2\leq n\leq 7$, and two if $n=8$ (since $(0,0,0,0,0,0,0,0,1)$ and $(1,1,1,1,1,1,1,1,3)$ leaves in different $\Gamma$-orbits).

What happens in the hermitian case?

More precisely, let $\Gamma=\mathrm{U}(n,1)\cap\mathrm M(n+1,\mathcal O)$ with $\mathcal O$ the ring of integer of a quadratic imaginary extension of $\mathbb Q$, and let $\mathcal R$ be the set of "integer" solutions $x\in\mathcal O^{n+1}$ of $$ Q[x]=x^*Qx=|x_1|^2+\dots+|x_{n}|^2-|x_{n+1}|^2=-1. $$

How many $\Gamma$-orbits are there in the set $\mathcal R$?

[Edit: It's enough for $n=2,3$ and some particular ring $\mathcal O$, like $\mathbb Z[\sqrt{-1}]$.]

Thank you in advance.-.