Assume we are given a complex vector space $V$ of dimension $n$ with a non-degenerate quadratic form $q$ (which is of course unique up to isomorphism).
If we take an element $v \in V$ such that $q(v) \neq 0$, then $W=v^\perp \subset V$ is an $n-1$-dimensional complex vector space and $q$ restricts to a non-degenerate quadratic form. The subgroup $Fix(v) \subset SO(V)$ is naturally isomorphic to $SO(W)$ in this case.
I am wondering what happens in the case that $q(v)=0$. I would like to better understand the (isomorphism type) of the (Lie) group $Fix(v) \subset SO(V)$.
Since any two complex quadratic forms are isomorphic over the complex numbers, after choosing a basis we know that $Fix(v)$ will be isomorphic to the group of special orthogonal transformations of a complex vector space $Z$ of dimension $n-1$ with a quadratic form with a one-dimensional radical. I.e. $Z$ has a basis $e_1, \dots, e_{n-1}$ and the quadratic form $q$ is of the form $q= x_2^2 + x_3^2 + \dots + x_{n-1}^2$.
Ultimately, I would like to understand the difference between $SO(W)$ from the first case and $SO(Z)$ from the second case. Are these Lie subgroups somehow related? Any help would be deeply appreciated.