Odd-dimensional rotations have an axis

Idea. The matrix $U-I$ is related to its transpose by a factor $-U$ up to conjugation; taking determinants gives an equation that forces $\det(U-I)=0$ when $n$ is odd.

Step 1: a factorisation. Since $UU^T=I$,
$$U-I=U-UU^T=U(I-U^T).$$

Step 2: determinants. Using $\det U=1$ and $\det M^T=\det M$,
$$\det(U-I)=\det U\cdot\det(I-U^T)=\det\bigl((I-U)^T\bigr)=\det(I-U).$$
But $I-U=-(U-I)$, so $\det(I-U)=(-1)^n\det(U-I)$. Therefore
$$\det(U-I)=(-1)^n\det(U-I).$$

Step 3: $n$ odd. For $n$ odd this reads $\det(U-I)=-\det(U-I)$, so $\det(U-I)=0$. Hence $1$ is an eigenvalue: there is a unit vector $e$ with $Ue=e$.

Euler's theorem. For $n=3$ this unit vector spans the axis, and $U$ restricts to an orientation preserving isometry of the plane $e^\perp$ (preserved because $U$ is orthogonal and fixes $e$), i.e. to a plane rotation. So $U$ is a rotation about $e$.

Remark. For $n$ even the statement fails: $\begin{pmatrix}0&-1\\1&0\end{pmatrix}\in SO(2)$ has no eigenvalue $1$. The same computation shows that for $n$ even and $\det U=-1$ one has $\det(U-I)=0$: every reflection-type orthogonal matrix has eigenvalue $1$ (and also $-1$).

Lesson. Parity of the dimension decides the existence of fixed vectors for rotations.