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The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$. When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.

There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$. The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.

In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$. One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line.

A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces. I would expect this example to be explicitly computed there.

Addendum: So, after thinking about it, I realized that the answer is this: Let $e_1,\ldots, e_n$ be a unitary basis of $\mathbb{C}^n$ with $e_1,e_2$ a basis of the fixed $\mathbb{C}^2$. Then each $2$-plane is in the $K$-orbit of a plane spanned by $$ \cos\theta_1\,e_1+\sin\theta_1\,e_2\quad\text{and}\quad \cos\theta_2\,e_3+\sin\theta_2\,e_4 $$ where $0\le \theta_1\le\theta_2\le\tfrac12\pi$, and the values $(\theta_1,\theta_2)$ in this triangle distinguish the orbits.

The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$. When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.

There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$. The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.

In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$. One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line.

A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces. I would expect this example to be explicitly computed there.

Addendum: So, after thinking about it, I realized that the answer is this: Let $e_1,\ldots, e_n$ be a unitary basis of $\mathbb{C}^n$ with $e_1,e_2$ a basis of the fixed $\mathbb{C}^2$. Then each $2$-plane is in the $K$-orbit of a plane spanned by $$ \cos\theta_1\,e_1+\sin\theta_1\,e_3\quad\text{and}\quad \cos\theta_2\,e_2+\sin\theta_2\,e_4 $$ where $0\le \theta_1\le\theta_2\le\tfrac12\pi$, and the values $(\theta_1,\theta_2)$ in this triangle distinguish the orbits.

The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$. When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.

There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$. The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.

In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$. One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line. (But don't trust my memory on that. It could be that one vertex is the set of $2$-planes that are graphs of Hermitian isometries from $\mathbb{C}^2$ to $\mathbb{C}^{n-2}$. It has been a while since I thought about this.)

A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces. I would expect this example to be explicitly computed there.

The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$. When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.

There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$. The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.

In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$. One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line.

A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces. I would expect this example to be explicitly computed there.

Addendum: So, after thinking about it, I realized that the answer is this: Let $e_1,\ldots, e_n$ be a unitary basis of $\mathbb{C}^n$ with $e_1,e_2$ a basis of the fixed $\mathbb{C}^2$. Then each $2$-plane is in the $K$-orbit of a plane spanned by $$ \cos\theta_1\,e_1+\sin\theta_1\,e_2\quad\text{and}\quad \cos\theta_2\,e_3+\sin\theta_2\,e_4 $$ where $0\le \theta_1\le\theta_2\le\tfrac12\pi$, and the values $(\theta_1,\theta_2)$ in this triangle distinguish the orbits.

The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$. When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.

There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$. The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.

In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$. One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line. (But don't trust my memory on that.)

A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces. I would expect this example to be explicitly computed there.

The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$. When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.

There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$. The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.

In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$. One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line. (But don't trust my memory on that. It could be that one vertex is the set of $2$-planes that are graphs of Hermitian isometries from $\mathbb{C}^2$ to $\mathbb{C}^{n-2}$. It has been a while since I thought about this.)

A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces. I would expect this example to be explicitly computed there.