= Solution
Taking <determinants> and using invariance under <transpose>,
$$
\det(I+A)=\det((I+A)^T)=\det(I-A),\qquad\boxed{\det N=1.}
$$
This <determinant> identity actually holds in every dimension. In dimension three, skew symmetry additionally gives
$$
\det A=\det A^T=\det(-A)=(-1)^3\det A=-\det A,
$$
so $\det A=0$. Hence there is a real nonzero <vector> $x$ in the <kernel>:
$$
\boxed{Ax=0,\qquad x\neq0.}
$$
The <eigenvalue> relation then gives $Nx=x$. Thus the <Cayley transform (Lie theory)> fixes an axis and is a proper three-dimensional rotation, including the identity as a degenerate case.
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