What are the solutions to $X$ for $X^{T} A X = A$? Knowing that, what are the solutions to Y for $Y= (I−X)(I+X)^{-1}$ (with $\det(I+X) \neq 0$)? All matrices are square matrices of real numbers. The goal is to use these properties to show that $$AY+Y^{T}A= 0$$
What do the equations in the title reveal about the properties of $A$, $X$, and $Y$? Can $X$ only be the identity matrix? 
Thanks!

loup blanc's edit: This question is about showing that the solutions of $X^TAX=A$ are in isomorphism with the solutions of $AY+Y^TA=0$. Recent research (2011) has made it possible to know the dimension of the solutions of the second equation, then the dimension of the solutions of the first one.
 A: In view of the expression for your $Y$, here is a pertinent result:

Theorem. Suppose $A$ is symmetric and nonsingular over a field of characteristic $\ne2$. Then every solution to the equation $X^TAX=A$ with $\det(I+X)\ne0$ is in the form of $X=(A+K)^{-1}(A-K)$ for some skew-symmetric matrix $K$ such that $A+K$ is nonsingular. Such a matrix $X$ is called a cogredient automorph or congruent automorph of $A$.

For a proof of the above theorem, see Sam Perlis, Theory of Matrices, pp.104-105.
In the special case where $A=I$ over $\mathbb R$, the equation $X^TAX=A$ reduces to $X^TX=I$ and hence $X$ is a real orthogonal matrix, and the expression $X=(A+K)^{-1}(A-K)$ in the theorem above becomes $X=(I+K)^{-1}(I-K)$, which is the familiar Cayley transform.
Some more general results for a nonsingular but perhaps non-symmetric $A$ can be found in sec. 37 (pp.65-68), chapter V of Mac Duffee's detailed survey The Theory of Matrices.
A: Without characterizing the solutions to your equations, we can say the following:
Suppose that $X$ is a solution to $X^TAX = A$, and that $Y = (I - X)(I + X)^{-1}$.  Because $\det(I+X) \neq 0$, it suffices to show that
$$
(I + X)^T[AY + Y^TA](I + X) = 0.
$$
With that in mind, we note that we can expand
$$
\begin{align*}
(I + X)^T[AY + Y^TA]&(I + X) =
(I+X)^TA(I-X) + (I-X)^TA(I + X) \\
&= A - AX + X^TA - X^TAX + A - X^TA + AX - X^TAX\\
&= A - AX + X^TA - A + A - X^TA + AX - A 
= 0.
\end{align*}
$$
The desired conclusion follows.
