= Solution
Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are <polynomials> in $B$. Also $P$ has a <matrix inverse> since it is the <transpose> of the invertible $Q$. Thus the <Cayley transform of a real skew-symmetric matrix>, with its parameter replaced by $-B$, satisfies
$$
C^TC=P^{-1}QQ^{-1}P=I.
$$
Hence $C$ is an <orthogonal matrix>. Its inverse is $Q P^{-1}$; this lets us distinguish the <symmetric and skew-symmetric Cayley transforms> below.
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