Let $(V,\omega)$ be a symplectic vector space. Let $W,W'$ be two isotropic subspaces. I want to see that there is symplectomorphism $\phi:V\rightarrow V$ such that $\phi(W)=W'$ if and only if $\dim(W)=\dim(W')$.
I know that every isotropic subspace is contained in a lagrangian subspace. I also know that if $L\subset V$ is a lagranian subspace, there is a symplectomorphism $W\times W^*\rightarrow V$
We have a nice result about lagrangian subspaces:
Let $L\subset V$ be a Lagrangian subspace. Consider $(L\times L^*,\Omega)$. Then there is a symplectomorphism $\Psi:L\times L^*\rightarrow V$ such that $\Psi(L\times\{0\})=L$.
It seems the idea is to show that for $W\subset L$ we have $\Psi(W\times\{0\})=W$. For $W'\subset L'$ we have $\Psi'(W'\times\{0\})=W'$.
Then by passing through $\mathbb R^{2n}$ we can get a symplectomorphism $L\times L^* \cong \mathbb R^{2n} \cong L'\times L'^*$.
I don't see why we have $\Psi(W\times\{0\})=W$ or why the symplectomorphism $L\times L^* \cong \mathbb R^{2n} \cong L'\times L'^*$ will send $W$ to $W'$
Let $k=\dim W=\dim W'\leq n$ where $\dim V=2n$.
You may use this fact
Symplectic basis $(A_i,B_i)$ such that $S= $ span$(A_1,...,A_k)$ for some $k$ when $S$ is isotropic
to give bases $A_1,...,A_k$ and $A_1',...,A_k'$ of $W$ and $W'$ respectively, where $A_1,...,A_n,B_1,...,B_n$ and $A'_1,...,A'_n,B'_1,...,B'_n$ are two symplectic bases of $V$, which therefore differ by a symplectic transformation.