jmart has modified the question since I posted my original answer. I'll now modify my answer to correspond:

(i) The Lagrangian Grassmannian ($L$ in your notation) is a closed submanifold of $Gr_n(\mathbb{R}^{2n})$, so it's not dense at all. In fact, it has dimension $\frac12n(n{+}1)$.

(ii) $L$ is homogeneous under the action of $\textrm{Sp}(n)\subset \textrm{GL}(2n,\mathbb{R})$, the subgroup that preserves the symplectic structure $\omega$. (The $J$ is not needed to define $L$.) Conversely, the subgroup of $\textrm{GL}(2n,\mathbb{R})$ that preserves $L$ in $Gr_n(\mathbb{R}^{2n})$ is easily seen to be the subgroup of $\textrm{GL}(2n,\mathbb{R})$ that preserves $\omega$ up to a multiple.

(iii) I don't really know what you mean by 'the orbits of $L$', since $L$ is not a group. Of course, $Gr_n(\mathbb{R}^{2n})$ is homogeneous under $\textrm{GL}(2n,\mathbb{R})$ so it's the $\textrm{GL}(2n,\mathbb{R})$-orbit of any point of $L$. Perhaps you actually want to know the orbits of $\textrm{Sp}(n)$ acting on $Gr_n(\mathbb{R}^{2n})$, with $L$ being the closed $\textrm{Sp}(n)$-orbit. If you consider, for any $n$-plane $E\in Gr_n(\mathbb{R}^{2n})$, the rank $r(E)$ of the pullback of $\omega$ to $E$, then the $\textrm{Sp}(n)$-orbits are the level sets of $r$ (which takes values in nonnegative integers).

(iv) I'm not sure what you mean by 'projections'. If by 'projection', you mean a smooth map $\pi:Gr_n(\mathbb{R}^{2n})\to L$ that is the identity on $L$, such a thing does not exist.