Euler's rotation theorem (very informally stated) says that an rotation of a sphere in three dimensional space always has two fixed points. I believe this claim generalizes to (n-1)-hyperspheres in n-dimensional space for odd n, but I am unable to patch the final gap in the proof. My strategy so far is as follows; any ideas on how to proceed would be greatly appreciated.
First, note that an arbitrary rotation of our hypersphere can be characterized by rotations about n-1 different axes. Since a rotation by theta in $\mathbb{R}^2$ is characterized by $\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{array}\right]$, it is not hard to see that a rotation about n-1 different axes can be characterized by the matrix product
$ \left[\begin{array}{ccccccc} \cos\theta_{1} & -\sin\theta_{1} & 0 & \cdots & & & 0\\ \sin\theta_{1} & \cos\theta_{1} & 0 & \cdots & & & 0\\ 0 & 0 & 1\\ \vdots & \vdots & & & & & \vdots\\ & & & & \ddots\\ & & & & & & 0\\ 0 & 0 & & \cdots & & 0 & 1\end{array}\right]\left[\begin{array}{ccccccc} 1 & 0 & 0 & \cdots & & & 0\\ 0 & \cos\theta_{2} & -\sin\theta_{2} & \cdots & & & 0\\ 0 & \sin\theta_{2} & \cos\theta_{2} & \cdots & & & 0\\ \vdots & \vdots & \vdots & 1 & & & \vdots\\ & & & & \ddots\\ & & & & & & 0\\ 0 & 0 & 0 & \cdots & & 0 & 1\end{array}\right] $
$ \cdots\left[\begin{array}{ccccccc} \cos\theta_{n-1} & 0 & & \cdots & & 0 & -\sin\theta_{n-1}\\ 0 & 1 & & \cdots & & & 0\\ \\\vdots & \vdots & & \ddots & & & \vdots\\ \\0 & & & & & 1 & 0\\ \sin\theta_{n-1} & 0 & & \cdots & & 0 & \cos\theta_{n-1}\end{array}\right] $.
Then notice that we have two fixed points if the matrix product just described has an eigenvalue of 1. The eigenvalues of a matrix are given by the roots of the characteristic polynomial; the characteristic polynomial here has real coefficents; therefore any complex roots of the characteristic polynomial must come in conjugate pairs; therefore if n is odd, we must have a real eigenvalue. But since rotations preserve length, any real eigenvalues must have an absolute value of one.
So we've shown that for odd n, there must be an eigenvalue of 1 or -1, but to complete the proof, we need to show that there can't be a lone eigenvalue of -1, and I haven't been able to figure out how to do that.
I thought I might be able to glean some intuition from the three-dimensional case, where a half-rotation about one axis gives us two eigenvalues of -1. Perhaps eigenvalues of -1 must also come in pairs? That seems plausible---dare I say even obvious?---on intuitive and physical grounds, but how might one prove it?
This excursion was inspired by an exercise in Bretscher Linear Algebra.