Symplectic transform between pairs of Lagrangian subspaces Let $V$ be a symplectic space. Suppose $(W_1, W_2)$ and $(U_1, U_2)$ are pairs of complementary Lagrangian subspaces. 
There exists a symplectic transform which maps $W_i$ into $U_i$, right? 
There are many such transforms actually, right? How to characterize them? 
 A: Such transformations always exist. We'll need some notation: Let $\Omega_0 = \left(\begin{smallmatrix} 0 & \operatorname{Id} \\ - \operatorname{Id} & 0 \end{smallmatrix}\right)$ be the standard symplectic form on $\mathbb{R}^{2n}$. 
Existence: We can use the symplectic Gram-Schmidt process to define a symplectomorphism $$\phi:(V,\Omega) \to (\mathbb{R}^{2n},\Omega_0)$$ that takes $W_1$ to $\mathbb{R}^n \times \{0\}^n$ and $W_2$ to $\{0\}^n \times \mathbb{R}^n$. Of course, we can find an analogous symplectomorphism $$\psi:(V,\Omega) \to (\mathbb{R}^{2n},\Omega_0)$$ taking $U_1$ to $\mathbb{R}^n \times \{0\}^n$ and $U_2$ to $\{0\}^n \times \mathbb{R}^n$. In this setup, we see that a symplectic map $V \to V$ taking $(W_1,W_2)$ to $(U_1,U_2)$ is equivalent to a symplectic map $A:\mathbb{R}^{2n} \to \mathbb{R}^{2n}$ which fixes $\mathbb{R}^n \times \{0\}^n$ and $\{0\}^n \times \mathbb{R}^n$. (Then $\psi^{-1} \circ A \circ \phi: V \to V$ is the originally sought-after map.) Since the identity $A=\operatorname{Id}_{2n}$  is one such transformation, the answer to your first question is yes, these maps exist. 
Classification: How many such maps are there?  First, the condition that $A$ fixes $\mathbb{R}^n \times \{0\}^n$ and $\{0\}^n \times \mathbb{R}^n$  means that it has the form
$$A= \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix},$$
for $A_1,A_2 \in GL(n,\mathbb{R})$. The symplectic condition $\Omega_0=A^T \Omega_0 A $ becomes
$$\begin{pmatrix} 0 & \operatorname{Id} \\ -\operatorname{Id} & 0 \end{pmatrix}=\begin{pmatrix} A_1^T & 0 \\ 0 & A_2^T \end{pmatrix}\begin{pmatrix} 0 & \operatorname{Id} \\ -\operatorname{Id} & 0 \end{pmatrix}\begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}=\begin{pmatrix} 0 & A_1^T A_2 \\ -A_2^T A_1 & 0 \end{pmatrix}.$$
This implies that we must have $A_2=(A_1^T)^{-1}$. So I believe the answer to your classification question is that the set of such maps $V \to V$ is one-to-one correspondence with the set of matrices
$$\begin{pmatrix} B & 0 \\ 0 & (B^T)^{-1} \end{pmatrix}$$
for all $B \in GL(n,\mathbb{R})$.
