= Solution
In odd dimension,
$$
\det A=\det A^T=\det(-A)=-\det A,
$$
so $\det A=0$. Since $A$ is real, its <kernel> contains a nonzero real vector $a$.
The plane $a^\perp$ is invariant under $A$, because $a\mathbin{\cdot}Ab=-(Aa)\mathbin{\cdot}b=0$. On this two-dimensional plane, $Ab$ is perpendicular to $b$. Put $|Ab|=\theta|b|$ with $\theta>0$. Then $A^2b$ is parallel to $b$, and
$$
b\mathbin{\cdot}A^2b=-(Ab)\mathbin{\cdot}(Ab)=-\theta^2|b|^2,
$$
so
$$
\boxed{A^2b=-\theta^2b}.
$$
The <matrix exponential> fixes the kernel vector:
$$
\boxed{e^Aa=a}.
$$
Separating even and odd powers in its series and using $A^{2j}b=(-1)^j\theta^{2j}b$ gives
$$
\boxed{e^Ab=\cos\theta\,b+\frac{\sin\theta}{\theta}Ab}.
$$
Thus $e^A$ acts as a rotation through angle $\theta$ on $a^\perp$ and fixes its axis $\mathbb Ra$.
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