Let $f\colon\mathbb{Z}^n\otimes\mathbb{Z}^n\to\mathbb{Z}$ be a non-degenerate symmetric bilinear form and consider the affine quadric hypersurface $$ X:=\{f(x,x)+2=0\}\subseteq\mathbb{Z}^n. $$ For each $x\in X$, there is a reflection $r_x\colon X\to X$ given by $r_x (y)=y+f(x,y)x$. Let $R$ be the subgroup of automorphisms of $X$ generated by all these reflections.
The question I want to ask is: does $R$ act transitively on $X$?
Not necessarily, not even if $f$ is negative definite: then $X$ is a Euclidean root system with all norms equal, and $R$ is its Weyl group, which is transitive iff the root system is irreducible.
The first counterexample appears for $n=2$ with $f$ being the standard negative-definite form $f(x,x') = -(x_1 x_1' + x_2 x_2')$. Here $X = \{(\pm1,\pm1)\}$. Taking $x = (1,1)$, we find for all $y = (y_1,y_2)$ that $$ r_x(y) = (y_1,y_2) - (y_1+y_2) (1,1) = (-y_2,-y_1), $$ which is always either $y$ or $-y$ for any $y \in X$, so there are two orbits $\{ \pm (1,1) \}$ and $\{ \pm (1,-1) \}$.
There are also more complicated examples with $f$ indefinite.
