Let $LG(n)$ be a Lagrangian Grassmanian manifold. That is, $LG(n)$ is the set of Lagrangian subspaces of a symplectic vector space of dimension $2n$.
Why can $LG(n)$ be identified with $U(n)/O(n)$?
Here $U(n)$ and $O(n)$ are unitary and orthogonal groups respectively.
The unitary group acts transitively on $\operatorname{LG}(n)$ i.e. there is a unitary action that can take any element of $\operatorname{LG}(n)$ to any other.
Considering a group $G$ acting on a manifold $X$ there is a homeomorphism: \begin{equation} f: \frac{G}{G_x} \leftrightarrow X \end{equation} where $G_x$ is the isotropy group of some $x \in X$.
For any $p \in \operatorname{LG}(n)$ the isotropy group is $\operatorname{O}(n)$. Therefore $\operatorname{LG}(n)$ is homeomorphic to $\operatorname{U}(n)/\operatorname{O}(n)$.
The general theorem is named Homogeneous Space Characterisation Theorem in Lee's 'Introduction to Smooth Manifolds'. The specific theorem for Lagrangian Grassmannians was found in a chapter of de Gosson's book, 'Symplectic Methods in Harmonic Analysis and in Mathematical Physics': webzoom.freewebs.com/cvdegosson/symplectic%20group.pdf