Define the unitary and Hermitian matrices
$$U=\left(
\begin{array}{cccc}
0 & 0 & -i & 0 \\
0 & 0 & 0 & -i \\
i & 0 & 0 & 0 \\
0 & i & 0 & 0 \\
\end{array}
\right),\;\;
V=\left(
\begin{array}{cccc}
0 & 0 & -i & 0 \\
0 & 0 & 0 & i \\
i & 0 & 0 & 0 \\
0 & -i & 0 & 0 \\
\end{array}
\right),
\;\;U^2=I=V^2,$$
and note that, for $\mu\in\mathbb{R}$,
$$UA(\mu)U=\bar{A}(\mu),\;\;VA(\mu)V=-A(\mu).$$

$\bullet$ Hence if $\lambda$ is an eigenvalue of $\prod_{i=1}^n A(\mu_i)$, then
$$0=\overline{\det\bigl(\lambda I-\prod_i A(\mu_i)\bigr)}=\det\bigl(\bar{\lambda} I-\prod_i\bar{A}(\mu_i)\bigr)=\det\bigl(\bar{\lambda} I-\prod_i UA(\mu_i)U\bigr)=\det\bigl(\bar{\lambda} I-\prod_i A(\mu_i)\bigr)=0.$$
So the eigenvalues come in complex conjugate pairs: if $\lambda$ is an eigenvalue of $\prod_i A(\mu_i)$, then also $\bar{\lambda}$ is an eigenvalue. (This holds irrespective of whether $n$ is even or odd.)

$\bullet$ Similarly, if $\lambda$ is an eigenvalue of $\prod_{i=1}^n A(\mu_i)$ and $n$ is an odd integer, then
$$0=\det\bigl(\lambda I-\prod_{i=1}^n VA(\mu_i)V\bigr)=\det\bigl(\lambda I-(-1)^n\prod_{i=1}^n A(\mu_i)\bigr)=\det\bigl(\lambda I+\prod_{i=1}^n A(\mu_i)\bigr)=0,$$
so the eigenvalues come in inverse pairs for odd $n$: if $\lambda$ is an eigenvalue then also $-\lambda$ is an eigenvalue. This proves property 2.

$\bullet$ Since $\det A(\mu)=1$ for any $\mu$, the product of the four eigenvalues of $\prod_{i=1}^n A(\mu_i)$ equals unity. This gives for odd $n$ the following three possibilities (with real $c$ and $\phi$):

**A.** $\lambda_1=ic$, $\lambda_2=-ic$, $\lambda_3=i/c$, $\lambda_4=-i/c$ (this is property 1),

**B.** $\lambda_1=c$, $\lambda_2=-c$, $\lambda_3=1/c$, $\lambda_4=-1/c$,

**C.** $\lambda_1=e^{i\phi}$, $\lambda_2=-e^{i\phi}$, $\lambda_3=e^{-i\phi}$, $\lambda_4=-e^{-i\phi}$.

The eigenvalues are either all four on the imaginary axis, or on the real axis, or on the unit circle.

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