Two equivalent matrices? By coincidence I noticed that the following two matrices yield the same eigenvalues
\begin{pmatrix} A & B \\ B^* & A \end{pmatrix} and \begin{pmatrix} 0& A+b1_{\mathbb C^{2 \times 2}} \\ A+b^* 1_{\mathbb C^{2 \times 2}} & 0 \end{pmatrix}
where $A = \begin{pmatrix} 0 & a \\ a^* & 0 \end{pmatrix}$ and $B=\begin{pmatrix} b & 0 \\ 0 & b^* \end{pmatrix}$ for any complex numbers, $a,b \in \mathbb{C}.$ Can one understand this somehow?

Define the unitary matrix
$$U=\left(
\begin{array}{cccc}
 i e^{i \pi /4} & 0 & 0 & 0 \\
 0 & 0 & 0 & -e^{-i \pi /4} \\
 0 & 0 & i e^{i \pi /4} & 0 \\
 0 & -e^{-i \pi /4} & 0 & 0 \\
\end{array}
\right),$$
then
$$U\begin{pmatrix} A & B \\ B^* & A \end{pmatrix}U^{-1}=\begin{pmatrix} 0 & A+b\mathbb{1} \\ A+b^\ast\mathbb{1} & 0 \end{pmatrix}$$
has the desired form.
Note also that the eigenvalues $\lambda$ of this matrix come in inverse pairs $\pm \lambda$, because it anticommutes with $\begin{pmatrix} \mathbb{1} & 0 \\ 0& -\mathbb{1} \end{pmatrix}$ (chiral symmetry).
