 Solution

Impose the discrete Z2 symmetry $\boxed{\Phi \mapsto -\Phi }$. The quadratic and quartic trace invariants are unchanged, while $\mathrm{tr}{\Phi }^3$ changes sign. Invariance therefore forbids a nonzero coefficient of that cubic operator. This is an additional symmetry of the scalar potential, not a consequence of SU(5) conjugation alone; cubic traces are permitted by the latter.