Let $(V,q)$ be a non-degenerate quadratic space. Then we know that for any $d$ with $0 \leq 2d \leq \dim V$, the group $O(q)$ of isometries of $(V,q)$ acts transitively on the set of totally isotropic subspaces of $V$ of dimension equal to $d$. (This is a consequence of a theorem of Witt.)
Is the same statement always true (maybe under some additional conditions) for $SO(q)$?
If the dimension of $V$ is even, $2n$, then there are two families of totally isotropic subspaces of dimension $n$ (if the form is split otherwise there may be no such subspace at all). Two such subspaces $W$ and $W'$ belong to the same family precisely when the parity of $\dim W\cap W'$ is the same as that of $n$. The two families are stable under $\mathrm{SO}(q)$ (and permuted by the rest of $\mathrm O(q)$). There certainly is an elementary proof of this fact yet I think that the "real" reason comes from algebraic geometry. The two families are disjoin algebraic subvarieties of the Grassmannian of $n$-spaces in $V$ and $\mathrm{SO}(q)$ is connected and hence must preserve them.
In any case when $n=1$ everything is very elementary; we can choose an isotropic basis $e_1,e_2$ with $e_1\cdot e_2=1$ and then $\mathrm{SO}(q)$ consists of the diagonal matrices of determinant $1$ which hence fixes the two isotropic subspaces (spanned by $e_1$ and $e_2$ respectively) while the orthogonal matrices of determinant $-1$ permute them.
In all other case $\mathrm{SO}(q)$ acts transitively on the totally isotropic spaces of fixed dimension (which is easily proven).
