Let $G=\operatorname{Sp}_{2r}(2)$. There are two orbits of $G$ on the natural $G$-set, one having the identity, the other having all the remaining elements. What is the subgroup of $G$ that stabilizes a nontrivial element of the $G$-set?
This might have no general answer or I'm being silly.. Please let me know.
On
It's a subgroup with structure $2^{2r-1}:{\rm Sp}(2(r-1),2)$, where the $2^{2r-1}$ denotes an elementary abelian group, and the extension is (I think) split.
The corresponding $(2r-1)$-dimensional module for ${\rm Sp}(2(r-1),2)$ over ${\mathbb F}_2$ is uniserial with a (minimal and maximal) submodule of dimension $1$.
It is a parabolic subgroup. (In symplectic groups, parabolic subgroups are stabilizers of totally isotropic subspaces.)
Say $v$ is a nonzero vector. Then you can extend this to a symplectic basis $v, e_1, ... e_{r-1}, f_{r-1}, ..., f_{1}, w$ where $\{v,w\}$ and $\{e_i, f_i\}$ are hyperbolic pairs.
Wrt. this basis, in the stabilizer of $v$ you have a Levi subgroup $$L = \{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & 1 \end{pmatrix} : A \in Sp(2r-2,2) \}$$
and a unipotent radical $$U = \{ \begin{pmatrix} 1 & * & * \\ 0 & I & * \\ 0 & 0 & 1 \end{pmatrix} \}$$
which has order $2^{2r-1}$. The stabilizer of $v$ in $Sp(2r,2)$ equals a semidirect product $U \rtimes L$.
So the structure is $2^{2r-1} : Sp(2r-2,2)$ as noted in the answer by Derek Holt.
Note that this also works for $r = 1$, in which case $L$ is trivial and $U$ is generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$.