Hermitian matrix has a symmetric spectrum if and only if $\lambda_1=\lambda_n$?

Question

I seem to remember a result claiming that an (irreducible) zero-diagonal $n\times n$ Hermitian matrix $H$ with entries $h_{ij}\in \{z\in\mathbb{C}~:~ |z|=1\}\cup\{0\}$ has eigenvalues $\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_n$ with the property that $\lambda_1=-\lambda_n$ if and only if $\lambda_{1+j}=-\lambda_{n-j}$ for all $j=0,1,\ldots,n-1$, but a source eludes me. Does this sound familiar to anyone or have I dreamt it?