Do all symplectic transformations give rise to skew symmetric matrices?

Question

Suppose that $ \Delta(x,y) = x^T\Delta y $ where $ \Delta$ is a symplectic matrix of form given in https://en.wikipedia.org/wiki/Symplectic_matrix

If I define an inner product $ \alpha(x,y) = \Delta(x,Ay) = x^T\alpha y $ (which implies that $ \alpha = \Delta A $)

then is it true that A is a symplectic matrix?

Using the fact that $ \Delta(x,y) = -\Delta(y,x) $ , I have reached the conclusion that

$ A^{T}\Delta = -\Delta A $

It is clear from this equation that $ A^T = -A $ satisfies the equation, though I am not sure that it is the only solution. Is there any other way to prove that A is indeed a skew-symmetric matrix? Assume all entries of all matrices to be real.

Refer the text in context from the book by Holevo on quantum information:

Answer 1

One should note, that $\Delta$ is the matrix representing the symplectic form on the vector space.

The matrix $A$ is symplectic if $$ A^T\Delta A=\Delta. $$ Now $A=\Delta^{-1}\alpha$. Thus using $\Delta^T = -\Delta = \Delta^{-1}$ $$ A^T\Delta A = (\alpha^T\Delta^{-T})\Delta \Delta^{-1}\alpha = \alpha\Delta\alpha. $$ In order that $A$ is symplectic, $\alpha$ has to be symplectic as well.

Does this make sense?