A nonsimply laced simple root system can be constructed from the simplylaced root system by folding the Dynkin diagram and hence the corresponding nonsimplylaced Lie algebra can be constructed by taking the fixed points of a nontrivial diagram automorphism (outer automorphism). Then how are their Grassmannians related ? Specifically, if $\sigma$ is an outer automorphism of $SL_{2n}$ and $P$ is a maximal parabolic, then is it true that $(G/P)^{\sigma}$ is a Grassamanian of $Sp_{2n}$ ? Is every Grassmannian of $Sp_{2n}$ obtained in this way ?
Most maximal parabolics of $SL_{2n}$ are not $\sigma$invariant, not even up to conjugation. So $(G/P)^\sigma$ does not make sense. The correct statement is: Let $I\subseteq\{1,\ldots,2n1\}$ be a symmetric subset, i.e., with $i\in I\Leftrightarrow 2ni\in I$. Let $P_I\subseteq SL_{2n}$ be the corresponding parabolic. Then $J:=I\cap\{1,\ldots,n\}$ corresponds to a parabolic $P_J\subseteq Sp_{2n}$. Then $\sigma(P_I)=P_I$ and $(SL_{2n}/P_I)^\sigma=Sp_{2n}/P_J$. In particular, the $i$th Grassmannian of $Sp_{2n}$ corresponds to the submaximal parabolic $P_I$ with $I=\{1,\ldots,2n1\}\setminus\{i,2ni\}$.
In geometric terms this corresponds to the following simple fact: the involution $\sigma$ induces an involution $U\mapsto U^\perp$ on subspaces of $\mathbb C^{2n}$. The $i$th Grassmannian of $Sp_{2n}$ consists of $i$dimensional isotropic subspaces $U$, i.e., with $U\subseteq U^\perp$. Thus, such a $U$ corresponds to a $\sigma$invariant partial flag $U\subseteq V$ with $\dim U=i$ and $\dim V=2ni$.

1$\begingroup$ If we take $I=\{n\}$ then $SL_{2n}/P_I=Gr(n,2n)$, where $P_I$ is the maximal parabolic corresponding to $I$. Then $(SL_{2n}/P_I)^{\sigma}=Sp_{2n}/P_n$ the Lagrangian Grassmannian. But the answer here suggests that the fixed point set is disconnected. mathoverflow.net/questions/266274/fixedpointsofaninvolution I am wondering what am I missing here ? $\endgroup$ – jack May 13 '17 at 3:03