Updated
Let $G:={\rm Aut}(X)$$G:={\rm Aut}(X)^0$. Then it is well-known that $G_{\mathbb C}={\rm Aut}(X_{\mathbb C})\cong {\rm PGL}(n,\mathbb C)$. $$G_{\mathbb C}= ({\rm Aut}(X)^0)_{\mathbb C}= ({\rm Aut}(X)_{\mathbb C})^0= {\rm Aut}(X_{\mathbb C})^0\cong {\rm PGL}(n,\mathbb C).$$ So $G$ is a real form of ${\rm PGL}(n,\mathbb C)$ which means that $G$ is isomorphic to ${\rm PGL}(n,\mathbb R)$, ${\rm PGL}(n/2,\mathbb H)$, or ${\rm PU}(p,n-p)$. On the other hand, $X$ has a real point $x$ which means $X\cong G/P$ with $P=G_x$. It follows from $G_{\mathbb C}/P_{\mathbb C}\cong X_{\mathbb C}\cong {\rm Gr}_{n,k,\mathbb C}$ that $P_{\mathbb C}$ is a maximal parabolic which is defined over $\mathbb R$. Looking at the Satake diagram of $G$ it corresponds therefore to a non-compact self-conjugate simple root $\alpha_k$. This gives the following possibilities $$ G={\rm PGL}(n,\mathbb R), k=1,\ldots,n-1\quad\Rightarrow\quad\text{your first case} $$ $$ G={\rm PGL}(n/2,\mathbb H), k=2,4\ldots,n-2\quad\Rightarrow\quad\text{your second case} $$ $$ G={\rm PU}(p,n-p), k=p=n/2\quad\Rightarrow\quad\text{your third case} $$ This argument should generalize to arbitrary fields. The first two cases combine to $X=$ space of $k$-dimensional subspaces of $D^n$ where $D$ is a central simple algebra.