Consider the general linera group $$ GL(n) = \left\lbrace \left(\begin{array}{cc} A & C \\ M & B \end{array}\right) \text{ with } A\in GL(k),\: B\in GL(n-k),\: M\in Mat(n-k,k),\: C\in Mat(k,n-k) \right\rbrace $$ of $n\times n$ invertible matrices, and the following $n\times n$ matrix $$ J = \left(\begin{array}{cc} I_{k} & 0 \\ 0 & 0 \end{array}\right) $$ where $I_k$ is the $k\times k$ identity matrix. If $M\in GL(n)$ then $$ MJM^T = \left(\begin{array}{cc} AA^T & AM^T \\ MA^T & MM^t \end{array}\right) $$ Now, consider the subgroup $F\subset G$ defined by imposing $M = 0$ and $AA^T = cI_k$ for some $c\neq 0$ in the base filed (that we can assume algebraically closed and of characteristic zero). Let $X_k = G/F$ be the orbit of $I_k$. Then $X_k$ has dimension $\frac{2nk-k^2+k-2}{2}$.

I would like to ask if anyone knows a method to compute the Picard group of $X_k$. This should be $\mathbb{Z}$ for $k = 1$ and $\mathbb{Z}/n\mathbb{Z}$ for $k = n$. Thank you.