Short version: I'm confused about maximal totally-isotropic flags versus maximal flags: do they have the same centralizer in the classical group?
Let $F$ be a field, $V$ be a finite dimensional vector space over $F$, and let $\beta:V \times V \to F$ be a bilinear form. Suppose $V$ has a basis $\{ e_1, e_2, \ldots, e_n, f_n, \ldots, f_2, f_1 \}$ so that $\beta(e_i,f_i) = \delta_{ij}$, $\beta(f_i,e_j)=-\delta_{ij}$, and $\beta(e_i,e_j)=\beta(f_i,f_j) = 0$ for all $i,j$.
Let $\mathcal{F}$ be the chain of subspaces $0 < \langle f_1 \rangle < \langle f_1, f_2 \rangle < \ldots < \langle f_1,f_2,\ldots, f_n \rangle$, and let $\mathcal{E}$ be the chain of spaces starting with $\mathcal{F}$ and continuing to $\ldots < \langle f_1, f_2, \ldots, f_n, e_n \rangle < \ldots < \langle f_1, f_2, \ldots, f_n, e_n, e_{n-1} \ldots, e_2, e_1 \rangle$.
Let $G = \{ g \in \operatorname{GL}(V) \mid \forall v,w \in V \beta( vg, wg ) = \beta(v,w) \}$ be the symplectic group of $V$, and let $U=C_G(\mathcal{F}) = \{ g \in G : \mathcal{F}_i (g_i - 1 ) \leq F_{i-1} \}$ be the centralizer of the flag, so that $g \in U$ iff $v_i \in \langle f_1, f_2, \ldots, f_i \rangle \implies v_i g = v_i + v_j$ for some $v_j \in \langle f_1, f_2, \ldots, f_{i-1} \rangle$.
Is $C_G(\mathcal{F}) \leq C_G(\mathcal{E})$?
I think this is equivalent to asking if $C_G(\mathcal{F})$ is a maximal unipotent subgroup (which it should be, I think).
I think it should have something to do with orthogonal decompositions of $V$, but when I try to write it out at the element level I don't get the right order of quantifiers to conclude $g$ centralizes $\mathcal{E}$.