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.