A map on Grassmannian Let $G=SL_{2n}$ and let $\sigma:G \to G$ be defined by $\sigma (A)= E(A^t)^{-1}E^{-1}$, where $E=antidiag(1,1, ... ,1,-1,-1,...,-1)$. Then the maximal parabolic associated to the simple root $\epsilon_n-\epsilon_{n+1}$ is $\sigma$-stable, so $\sigma$ induces a map on $Gr(n,2n)$. The symplectic form associated to $E$ is $(X,Y) \mapsto X^tEY$. Is the induced map of $\sigma$ on $Gr(n,2n)$ given by $W \mapsto W^{\perp}$ ?
 A: I am just posting my comment as an answer.  I will change slightly your definition of $E$ so that the associated bilinear skew-symmetric form is the "standard" form, i.e., $$[ (x_1,\dots,x_{2n}), (y_1,\dots,y_{2n}) ] = (x_1y_{n+1}-x_{n+1}y_1) + \dots + (x_ny_{2n}-x_{2n}y_n).$$  For this symplectic form, the matrix $E$ has the following $(n|n)$-block form, $$ E=\left[\begin{array}{c|c} 0_{n\times n} & \text{Id}_{n\times n} \\ \hline -\text{Id}_{n\times n} & 0_{n\times n} \end{array} \right]. $$ The maximal parabolic subgroup $P$ preserved by the involution is the group of all matrices of $(n|n)$-block form,
$$
R=\left[\begin{array}{c|c} A & B \\ \hline 0_{n\times n} & D \end{array} \right].
$$
The involute of this matrix is
$$
\sigma(R) = \left[\begin{array}{c|c} (D^\dagger)^{-1} & B' \\ \hline 0_{n\times n} & (A^\dagger)^{-1} \end{array} \right],
$$
where $B'$ equals $(D^\dagger)^{-1}B^\dagger (A^\dagger)^{-1}$.
The open Bruhat cell in $\textbf{SL}_{2n}/P$ is the set of cosets $M\cdot P$ for $M$ of the following form,
$$
M=\left[\begin{array}{c|c} \text{Id}_{n\times n} & 0_{n\times n} \\ \hline C & \text{Id}_{n\times n} \end{array} \right].
$$ 
For such $M$, the involute $\sigma(M)$ equals,
$$
N=\sigma(M) = \left[\begin{array}{c|c} \text{Id}_{n\times n} & 0_{n\times n} \\ \hline C^\dagger & \text{Id}_{n\times n} \end{array} \right].
$$
The pairings of the first $n$ column vectors of $M$ and $N$ can be computed via the matrix product,
$$
N^\dagger E M = \left[\begin{array}{c|c} \text{Id}_{n\times n} & C \\ \hline 0_{n\times n} & \text{Id}_{n\times n} \end{array} \right]
\left[\begin{array}{c|c} 0_{n\times n} & \text{Id}_{n\times n} \\ \hline -\text{Id}_{n\times n} & 0_{n\times n}\end{array} \right]
\left[\begin{array}{c|c} \text{Id}_{n\times n} & 0_{n\times n} \\ \hline C & \text{Id}_{n\times n} \end{array} \right].
$$
This product equals $E$.  The zero block in the upper left means that the first $n$ column vectors of $M$ and $N$ are orthogonal with respect to your symplectic form.  Thus, on the open Bruhat cell $\sigma$ sends each $n$-dimensional subspace to its orthogonal.  Since the open Bruhat cell is Zariski dense, this holds on the entire Grassmannian.
