Let $G=\operatorname{SO}^{+}_{2n}(2)$. I did some Magma computation and found there were $3$ orbits on the natural $G$-set when $n=2,3,4$. The orbit sizes are $1$-$9$-$6$, $1$-$35$-$28$, $1$-$135$-$120$.
Is there a formula for the orbit sizes in general? Thank you.
As Mikko Korhonen said in the comments, the (non-zero) singular and non-singular vectors form single orbits by Witt's lemma.
So it remains to count the singular vectors, which are (row) vectors ${\mathbf v}$ over ${\mathbb F}_2$ such that ${\mathbf v}M {\mathbf v}^{\mathsf T} = 0$, where $M$ is the degree $2n$ matrix $$\left(\begin{array}{cc}{\mathbf 0}&{\mathbf E}\\{\mathbf 0}&{\mathbf 0}\end{array}\right)\quad{\rm with}\quad {\mathbf E}=\left(\begin{array}{ccccc}0&0&\cdots&0&1\\0&0&\cdots&1&0\\&&\cdots&&\\0&1&\cdots&0&0\\1&0&\cdots&0&0\end{array}\right).$$ So we have to count the number of vectors ${\mathbf v}=(x_1,x_2,\ldots,x_{2n})$ with $$x_nx_{n+1}+x_{n-1}x_{n+2}+\cdots x_1x_{2n}=0.$$ This is a straightforward induction. For the base case $m=1$ there are three pairs $x_1,x_2$ with $x_1x_2=0$. For the inductive step, we have $2^{2n-3} + 2^{n-2}$ choices for $x_1,\ldots,x_{n-1}x_{n+2},\ldots,x_{2n}$ with $x_{n-1}x_{n+2}+\cdots x_1x_{2n}=0$ and the number of of choices for $x_1,x_2,\ldots,x_{2n}$ is $$3(2^{2n-3} + 2^{n-2}) + (2^{2n-3} - 2^{n-2}) = 2^{2n-1} + 2^{n-1}.$$
