Is it always true that $Mat(id, F, E)=Mat(id, E,F)^{-1}$? Is it always true for change of basis matrices from and to any bases $E,F \subset \mathbb{R}^n$ that 
$$Mat(id, F, E)=Mat(id,E,F)^{-1}$$?

$Mat(id,F,E)$ is a change of basis matrix from basis $E \subset \mathbb{R}^n$ to $F \subset \mathbb{R}^n$.
 A: The way you phrase the problem is incorrect. When proving such equalities regarding matrices, you must have an ordered basis. Otherwise, the equalities would obviously be false. 
So let's fix two ordered bases $\alpha$ and $\beta$ for any finite dimensional vector space $V$. Let $T$ be a linear operator on $V$. I shall use the following notation:
$[T]_\alpha ^\beta$ to mean the matrix representation of $T$ in the ordered bases of $\alpha$ and $\beta$. Then the change of coordinate matrix can be represented by $[I]_\alpha ^\beta$

Proposition: For a linear operator $T$ on a finite dimensional vector space $V$ with ordered bases $\alpha$ and $\beta$, $T$ is invertible if and only if $[T]_\alpha ^ \beta$ is invertible. Moreover, $[T^{-1}]_\beta ^ \alpha = ([T]_\alpha ^ \beta ) ^{-1}$

This is easily verified. For example, if $T$ is invertible linear operator, then there exists $T^{-1}$. Then
$$I = [I]_\alpha^\alpha= [T^{-1} T]_\alpha ^ \alpha = [T^{-1}]_\beta ^\alpha [T]_\alpha ^ \beta$$
Specializing to $T = I$, which is clearly invertible, then you have that the inverse to $[I]_\alpha ^ \beta$ is given by $[I^{-1}]_\beta ^ \alpha = [I]_\beta ^ \alpha$. 
