Consider a real vector space $T$ of dimension $p+q$ with a non-degenerate symmetric bilinear form, $B:T\times T\to\mathbb{R}$, with signature $(p,q)$. Define the cone $$ C=\{(x_1,\dots,x_p,x_{p+1},\dots,x_{p+q})\in T\setminus\{0\}: x_1^2+\cdots+x_p^2-x_{p+1}^2-\cdots-x_{p+q}^2=0\}, $$ which is nothing else than the set of isotropic vectors w.r.t. the bilinear form.
I know that if $p=0$ or $q=0$ the cone is empty, so we are considering the cases where $\min\{p,q\}\geq 1$. I also know that this is false for $p=q=1$, while the action of $O(1,1)$ is transitive: an explicit computation of $SO(1,1)$ and $C$ easily proves it.
But this must be true for $\min\{p,q\}\geq 1$ and $\max\{p,q\}>1$.
Given two vectors $v,w\in C$ I think the proof should be divided in three different cases:
$v$ and $w$ are linearly dependent: In this case we can find a two dimensional subspace where the bilinear form has signature $(1,1)$ and choose a basis for which $v=(1,1,0\dots,0)$. In this case we can find an element in $SO(1,1)$ that contracts or expands $v$ to $w$ (even if there is a change of sign).
$v$ and $w$ are linearly dependent and $B(v,w)=0$. In this case we can find a vector $u\in C$ such that $B(v,u)\neq 0$ and $B(w,u)\neq 0$. Then there are two transformations in $O(1,1)\setminus SO(1,1)$ that take $v$ to $u$ and $u$ to $w$, respectively. Taking the determinant of their composition shows it is in $SO(1,1)$.
$v$ and $w$ are linearly independent and B$(v,w)\neq 0$. This is where I'm stuck.
I tried showing it for $SO(2,1)$ and $SO(1,2)$ explicitely, but calculating the explicit matrix form of their elements is quite cumbersome. Any other proof or idea is welcome.
I would also appreciate if someone could tell me if what I've done is correct.