= Solution
Multiplying the defining equation on the right by $I+A$ gives $(I+U)A=I-U$. Consequently the only possible answer is
$$
\boxed{A=(I+U)^{-1}(I-U)=(I-U)(I+U)^{-1}.}
$$
The factors commute because they are rational functions of $U$. Its <conjugate transpose>, using the <unitarity> $U^\dagger=U^{-1}$, is
$$
\begin{aligned}
A^\dagger&=(I+U^{-1})^{-1}(I-U^{-1})\\
&=(I+U)^{-1}U\bigl[-U^{-1}(I-U)\bigr]=-A.
\end{aligned}
$$
Thus $A$ is <Skew-Hermitian>. Moreover $I+A=2(I+U)^{-1}$ and $I-A=2U(I+U)^{-1}$, so substitution recovers $U$ and establishes existence as well as uniqueness. The <unitary matrix> $U=-I$ is a counterexample to invertibility of $I+U$; more generally the obstruction is an <eigenvalue> $-1$.
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