Let $V=\mathbb{R}^4$. Let $S$ be the set of all two-dimensional subspaces of $V$ and fix $W\in S$. Let $G=GL(V)$ (the group of invertible linear operators on $V$) act naturally on $S$ and let $H=\{g\in G\,\vert\,gW=W\}.$ Show that $H$ has exactly 3 orbits on $S$.
Attempt: I know that $G$ acts transitively on $S$, so $G$ only has one orbit in $S$. When we restrict the action to $H$, then $W$ is in its own orbit, and so $H$ has at least two orbits on $S$.
I was thinking of looking at the matrix representation of $H$ and trying to calculate the orbits of $H$ on $S$ directly, but that seems quite laborious. Is there a better way to approach this problem?