= Solution
Use $U^\dagger=U^{-1}$ and take the <Hermitian conjugate> in the reverse product order:
$$
B^\dagger=-i(I-U^{-1})^{-1}(I+U^{-1}).
$$
The factorization $I-U^{-1}=-U^{-1}(I-U)$ gives $(I-U^{-1})^{-1}=-(I-U)^{-1}U$, while $I+U^{-1}=U^{-1}(I+U)$. Consequently
$$
B^\dagger=i(I-U)^{-1}(I+U)=i(I+U)(I-U)^{-1}=\boxed{B}.
$$
The last equality uses the fact that $I+U$ commutes with $I-U$ and its inverse. Thus $B$ is a <Hermitian matrix>. The exclusion of the <eigenvalue> one is essential to define the inverse. Spectrally, an <eigenvalue> $e^{i\vartheta}\ne1$ is sent to $-\cot(\vartheta/2)$, a real number. This is the <inverse Cayley transform of a unitary matrix without eigenvalue one>.
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