 Solution

The spectral theorem for a Hermitian matrix gives a unitary diagonalizer $V$ with $V\Phi V^{†}=\mathrm{diag}(x_1,\dots ,x_5)$ and real $x_i$. Similarity preserves the trace, so ${\sum }_i{x}_i=0$.

To ensure that the diagonalizer belongs to SU(5) rather than only the unitary group $U(5)$, write $\det V=e^{i\alpha }$ and put $U=e^{-i\alpha /5}V$. Its determinant is one, and the scalar phase cancels from its conjugation action. ThusThe same argument shows why no additional reality condition beyond Hermiticity is needed. Permuting the real eigenvalues merely changes their order in the diagonal form.