= Solution
\b[Commuting inverse.] Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication gives
$$
(I+M)(I-M)=I-M^2=(I-M)(I+M).
$$
Multiply on both sides by $(I-M)^{-1}$ to obtain
$$
\boxed{(I+M)(I-M)^{-1}=(I-M)^{-1}(I+M).}
$$
More generally, a <matrix> commuting with an invertible <matrix> also commutes with its inverse.
\b[Orthogonality.] For a real <skew-symmetric matrix> $A$, $x^TAx=0$. If $(I-A)x=0$ for a real <vector> $x$, multiply by $x^T$ to obtain $\|x\|^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the <Cayley transform of a real skew-symmetric matrix>,
$$
N=(I+A)(I-A)^{-1},\qquad N^T=(I+A)^{-1}(I-A).
$$
The factors commute by the first result, so
$$
N^TN=(I+A)^{-1}(I-A)(I+A)(I-A)^{-1}=I.
$$
Therefore $\boxed{N\text{ is an orthogonal matrix}.}$
\b[Eigenvector mapping.] If $Av=av$, then $(I-A)^{-1}v=(1-a)^{-1}v$. Consequently
$$
\boxed{Nv=\frac{1+a}{1-a}v.}
$$
The denominator cannot vanish. In fact the <eigenvalues> of a real <skew-symmetric matrix> are purely imaginary: $v^\dagger Av$ is the negative of its complex conjugate, so $a=-\overline a$. For $a=i\beta$ the corresponding <eigenvalue> of $N$ is $(1+i\beta)/(1-i\beta)=e^{2i\arctan\beta}$, consistent with <orthogonality>.
\b[Spectrum in three dimensions.] Let $Q$ be any real three-by-three <orthogonal matrix>. For a complex <eigenvector> $v$, $\|Qv\|=\|v\|$ implies $|\lambda|=1$. A real cubic <characteristic polynomial> has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the <eigenvalues> can be written
$$
\boxed{e^{i\alpha},\quad e^{-i\alpha},\quad\varepsilon,\qquad\alpha\in\mathbb R,\quad\varepsilon\in\{1,-1\}.}
$$
Their product is $\varepsilon=\det Q$.
\b[Determinant and fixed axis for the Cayley transform.] Since $A^T=-A$,
$$
\det(I+A)=\det((I+A)^T)=\det(I-A),
$$
so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an <eigenvector> with $Nx=x$. Set $z=(I-A)^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. Hence
$$
\boxed{Ax=0\text{ for a nonzero real direction }x.}
$$
This is the axis fixed by the three-dimensional <rotation>. Equivalently, a real odd-dimensional <skew-symmetric matrix> has zero <determinant>, because $\det A=\det A^T=\det(-A)=-\det A$.
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