**Answer:** ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$; see Corollary 4 below.

**Theorem 1.** *Let $G$ be a simply connected semisimple group over a field $k$ of characteristic 0.
Let $H\subset G$ be an algebraic subgroup defined over $k$, not necessarily connected. Set $X=G/H$.
Then there is a canonical isomorphism ${\rm Pic\,} X={\widehat H}(k)$, where ${\widehat H}(k) ={\rm Hom}_k(H,{\Bbb G}_{m})$
is the character group of $H$.*

*Proof.* First assume that $H$ is connected. We deduce the theorem from results of the paper
J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.
J. Reine Angew. Math. 327 (1981), 12–80.

By Proposition 6.10 of this paper, there is a natural exact sequence of abelian groups
$${\widehat G}(k)\to {\widehat H}(k)\to{\rm Pic\,} X\to {\rm Pic\,} G.$$
Clearly we have ${\widehat G}(k)=0$. By Sansuc's Lemma 6.9(iv), we have ${\rm Pic\,} G=0$ (here Sansuc refers to a paper by Fossum and Iversen).
We obtain an isomorphism ${\widehat H}(k)= {\rm Pic\,} X$, as required.

Now we do not assume that $H$ is connected. We deduce Theorem 1 from a general result of
M. Borovoi and J. van Hamel, Extended equivariant Picard complexes and homogeneous spaces. Transform. Groups 17 (2012), 51-86.
Since ${\rm Pic\,} G_{\bar k}=0$ and $X$ has $k$-points, by Theorem 2 (Theorem 7.1) of this paper there is a canonical isomorphism
$$ {\rm Pic\,} X=H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle).$$
Here ${\bar k}$ is an algebraic closure of $k$, ${\widehat H}({\bar k})={\rm Hom}_{\bar k}(H,{\Bbb G}_{m})$, and similarly for ${\widehat G}({\bar k})$.
Further, $[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle$ denotes the complex of ${\rm Gal}({\bar k}/k)$-modules
$$\dots \to 0\to {\widehat G}({\bar k})\to {\widehat H}({\bar k})\to 0\to \dots$$
with ${\widehat H}({\bar k})$ in degree 1, and $H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle)$ denotes the first Galois hypercohomology of this complex.

In our case ${\widehat G}({\bar k})=0$, and therefore,
$$ {\rm Pic\,} X=H^1(k,[0\to {\widehat H}({\bar k})\rangle)=H^0(k,{\widehat H}({\bar k}))={\widehat H}(k),$$
as required.

This looks like killing a fly with a bazooka, and there should be an elementary proof of Theorem 1.

**Construction 2.**
The class in ${\rm Pic\,} X$ corresponding to a character
$$\chi\colon H\to{\Bbb G}_m$$
is described as follows. We consider the direct product $G\times {\Bbb G}_m$ and the injective homomorphism
$$\iota_\chi\colon H\to G\times {\Bbb G}_m,\quad h\mapsto (h,\chi(h)).$$
Further, we consider the quotient $Y_\chi:=(G\times {\Bbb G}_m)/\iota_\chi(H)$ and the projection map
\begin{gather*}\pi\colon\, Y_\chi=(G\times {\Bbb G}_m)/\iota_\chi(H)\,\longrightarrow\, G/H=X,\quad \\
[g,c]\,\mapsto\, [g]\quad \text{for }g\in G,\ c\in{\Bbb C}^\times.\end{gather*}
The group ${\Bbb G}_m$ acts on the fibers of $\pi$ by $c'\cdot [g,c]=[g,c'c]$ for $c'\in{\Bbb C}^\times$.
We see that $\pi\colon Y_\chi\to X$ is a principal ${\Bbb G}_m$-bundle over $X$.
To $\chi$ we associate the class of $Y_\chi$ in ${\rm Pic\,} X$.

We compute the character group $\widehat H$ of the stabilizer $H={\rm Sp}(2n)\cap{\rm GO}(2n)$, where
$$ {\rm GO}(2n)=\{A\in{\rm GL}(2n,{\Bbb C})\mid A^t A=\lambda_A I,\ \lambda_A\in{\Bbb C}^\times\}.$$

**Proposition 3.** *For $H={\rm Sp}(2n)\cap{\rm GO}(2n)$
we have* ${\widehat H}={\Bbb Z}/2{\Bbb Z}$.

*Proof.*
We compute the group $H$.
We write the equations for $A\in H$:
$$
A^t A =\lambda_A I,\qquad A^t J A=J, \qquad\text{where } J=
\begin{pmatrix} 0 & I_n\\ -I_n &0 \end{pmatrix}.
$$
We obtain
$$\lambda_A A^{-1} J A=J, \quad\text{whence } \lambda_A J A=AJ.$$
Let $x$ be an eigenvector of $J$ with eigenvalue $\mu$.
Then
$$ Jx=\mu x,$$
whence
$$AJx=\mu Ax,\qquad \lambda_A JAx=\mu Ax,\qquad Jy=\lambda_A^{-1} \mu y, \text{ where }y=Ax.$$
We see that $y$ is an eigenvector of the matrix $J$ with eigenvalue $\lambda_A^{-1}$.
Thus $\lambda_A^{-1}\mu$ is an eigenvalue of $J$ as well.
Since our matrix $J$ has only two eigenvalues $i$ and $-i$, we conclude that $\lambda_A$ can take values only $1$ and $-1$. Thus we obtain a homomorphism
$$\lambda\colon H\to \mu_2,\quad A\mapsto \lambda_A.$$

Consider the matrix
$$ S=i\begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}. $$
An easy calsulation shows that

$$ S^t S=S^2=-I,\qquad S^t J S=SJS=J.$$
Thus $S\in H$, $\lambda_S=-1$.
We obtain a short exact sequence
$$ 1\to H_1\to H\to \mu_2\to 1,$$
where $H_1={\rm Sp}(2n)\cap{\rm SO}(2,n)$ and where
the homomorphism $\lambda\colon H\to\mu_2$ is surjective because $\lambda_S=-1$.
We have $H=H_1\cup S\cdot H_1$.

The group $H_1$ was computed by Sasha in his answer:
it is isomorphic to ${\rm GL}(n,{\Bbb C})$ acting on $V=L_1\oplus L_2$ by
$B\mapsto (B,B^{-1})$. The linear operator $S$ permutes the subspaces $L_1$ and $L_2$, and it acts on the normal subgroup $H_1$ of $H$ as follows:
$$ S\cdot (B,B^{-1}) \cdot S^{-1}=(B^{-1},B).$$
Hence
$$ S\cdot (B,B^{-1}) \cdot S^{-1}\cdot (B,B^{-1})^{-1}=(B^{-2},B^2).$$
It follows that the commutator subgroup $(H,H)$ of $H$ is $H_1$.
Thus
$${\widehat H}=\widehat{H/H_1}=\widehat{\mu_2}={\Bbb Z}/2{\Bbb Z},$$
as required. The nontrivial element of the character group ${\widehat H}$ is the character
$$\lambda\colon H\to \mu_2\hookrightarrow{\Bbb G}_m,\quad
A\mapsto \lambda_A\in {\Bbb C}^\times.$$

**Corollary 4.** *For $X={\rm Sp}(2n)/({\rm Sp}(2n)\cap {\rm GO}(2n))$ we have* ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$.

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