Let $\Xi$ be the set of all real $n\times n$ which do not have $1$ as an eigenvalue. Consider the mapping
\begin{equation}\phi:\Xi\rightarrow\mathbb{R}^{n\times n},\quad \phi(A)=(A-I)^{-1}(A+I).\end{equation}
Prove that $\phi(A)\in\Xi$ and that $A$ is skew-symmetric if and only if $\phi(A)$ is orthogonal.
I have managed to show the later part of the question as follows:
$(\implies)$Let $A$ be skew-symmetric: \begin{align}\phi(A)\phi(A)^T&=(A-I)^{-1}(A+I)(-A+I)(-A-I)^{-1}\\ &=(A+I)(A-I)^{-1}(-A+I)(-A-I)^{-1}\\ &=-(A+I)(A-I)^{-1}(A-I)(-A-I)^{-1}\\ &=-(A+I)(-A-I)^{-1}\\ &=(-A-I)(-A-I)^{-1}=I\\\end{align} Hence $\phi(A)$ is orthogonal.
$(\Longleftarrow)$ Let $\phi(A)$ be orthogonal: \begin{align}\phi(A)\phi(A)^T&=I\\ (A-I)^{-1}(A+I)[(A-I)&^{-1}(A+I)]^T=I\\ (A-I)^{-1}(A+I)(A+I)&^{T}((A-I)^{-1})^T=I\\ (A-I)^{-1}(A+I)(A^T+&I)(A^T-I)^{-1}=I\\ (A+I)(A^T+I)(A^T&-I)^{-1}=(A-I)\\ (A+I)(A^T+I)=&(A-I)(A^T-I)\end{align} We know that we require this last line to commute, and we can see that to have these commute we require that $A^T=-A,$ i.e. we require $A$ skew-symmetric: \begin{align}(A+I)(-A+I)&=(A-I)(-A-I)\\ -(A+I)(A-I)&=-(A-I)(A+I)\\ (A+I)(A-I)&=(A-I)(A+I).\end{align} Therefore $A$ is skew-symmetric $\iff$ $\phi(A)$ is orthogonal.
I am pretty confident that my $\implies$ is correct, but would like a sanity check on the proof for $\Longleftarrow$.
As for showing that $\phi(A)\in\Xi,$ I am unsure how to tackle this.
We compute: $$ \begin{aligned} \phi(A)-I &=(A-I)^{-1}(A+I)-I\\ &=(A-I)^{-1}[\ (A+I)-(A-I)]\\ &=2(A-I)^{-1}\ , \\[3mm] &\qquad\text{ so} \\ \det(\phi(A)-I) &=2^n\det(\ (A-I)^{-1}\ ) \\ &\ne 0\ , \end{aligned} $$ (in characteristic $\ne 2$, yes, we are working over reals,) so $1$ is not an eigenvalue.
The first implication is ok.
For the second implication ($\Leftarrow$), just open the parenteses in $(A+I)(A^T+I)=(A-I)(A^T-I)$ when reaching this point to get $2(A+A^T)=0$.