= Solution
For a possibly complex <eigenvector> $v\neq0$ with $Av=\lambda v$, invertibility of $I-A$ gives $(I-A)^{-1}v=(1-\lambda)^{-1}v$. Therefore
$$
\boxed{Nv=\frac{1+\lambda}{1-\lambda}v,\qquad\nu=\frac{1+\lambda}{1-\lambda}.}
$$
Thus every <eigenvector> of $A$ remains an <eigenvector> of $N$. For a real <skew-symmetric matrix>, its <eigenvalues> have the form $\lambda=it$ with $t$ real, so $\nu=e^{2i\arctan t}$ lies on the <unit circle>, consistently with <orthogonality>.
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