The statement is the following:
Let $V_1$ and $V_2$ be two symplectic vector spaces with the same dimension and $F_i\subset (V_i,\omega_i),i=1,2$ a subspace of $V_i$. Assume that there exists a linear isomorphism $\phi:F_1\cong F_2$, i.e., $\phi^*(\omega_2|F_2)=\omega_1|F_1$. Then $\phi$ extends to a symplectic map $\tilde{\phi}:(V_1,\omega_1)\rightarrow (V_2,\omega_2)$.
$\bf{Hint}$: show that symplectic maps are the same thing as Lagrangians in $(V_1\oplus V_2,\omega_1-\omega_2)$ which are transverse to $V_1$ and $V_2$, and the map we are looking for, correspond to Lagrangians transverse to $V_1$ and $V_2$ containing $I=\{(x,\phi(x))|x\in F_1\}$. Compute the dimension of the non transverse ones.
Following the hint, I proved that there is a 1-1 correspondence between the set of symplectic maps from $(V_1,\omega_1)$ to $(V_2,\omega_2)$ and the set of Lagrangians in $(V_1\oplus V_2,\omega_1-\omega_2)$ transverse to $V_1$ and $V_2$. And then I checked that $I$ is an isotropic subspace of $V_1\oplus V_2$. So the remaining is proving that there exists a Lagrangian subspace $L$ containing $I$ and transverse to $V_1$ and $V_2$. I think I need to compute the dimension of the non transverse ones written in the hint. But I don't know how to compute the dimension and to prove the existence of such Lagrangian.
If someone knows, please help me.