= Solution
Since both real <matrices> are antisymmetric, $A^\dagger=-A$ and $B^\dagger=-B$. Thus $U^\dagger=-A+iB$. Expand the two unitary identities:
$$
UU^\dagger=-A^2-B^2+i(AB-BA)=I,\qquad U^\dagger U=-A^2-B^2-i(AB-BA)=I.
$$
Subtracting and adding give $\boxed{AB=BA,\qquad A^2+B^2=-I}$. Because $A$ has simple <eigenvalues>, each of its complex <eigenspaces> is one-dimensional. Commutation makes each one invariant under $B$: if $Av=av$, then $A(Bv)=a(Bv)$. Hence $Bv=bv$ for some scalar $b$. The complete <eigenbasis> of $A$ is therefore a common <eigenbasis>, which also diagonalizes $U$.
The <skew-Hermitian decomposition of a unitary matrix> is explicit:
$$
A=\frac{U-U^\dagger}{2},\qquad B=\frac{U+U^\dagger}{2i}.
$$
If $Uv=uv$, unitarity gives $U^\dagger v=u^{-1}v=\overline u\,v$. Consequently
$$
\boxed{a=\frac{u-\overline u}{2}=i\operatorname{Im}u,\qquad b=\frac{u+\overline u}{2i}=-i\operatorname{Re}u.}
$$
Equivalently, for $u=e^{i\theta}$ the paired <eigenvalues> are $\boxed{a=i\sin\theta,\ b=-i\cos\theta}$. The hypotheses can only occur in even dimension: $U^T=-U$ and $U$ is invertible, whereas an odd-dimensional <skew-symmetric matrix> has zero <determinant>.
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