Is $\Phi_5(z) \Phi_6(z) = 1 + z^2 + z^3 + z^4 + z^6$ the only product of cyclotomic polynomials that has nonnegative coefficients and satisfies $p(\zeta)=0$, $p(\zeta^2)=2$, $p(\zeta^3)=3$, and $p(1)=5$ where $\zeta$ is a primitive 6th root of unity?
This question arises from my continuing efforts to understand the cyclic sieving phenomenon (see Evaluating products of cyclotomic polynomials at roots of unity). Orbit-size data come nowhere close to determining the sieving polynomial $p(z)$, but I suspect that imposing extra constraints on the polynomial (specifically that $p(z)$ is a product of cyclotomic polynomials and that all coefficients are nonnegative) might do the trick (or at least reduce the set of candidates to a finite set).
This special case seems like a good place to start (though I'd be interested in other cases as well).
There are plenty of cyclotomic polynomials $q(z)$ that take the value 1 at $\zeta^2$ and $\zeta^3$ and 1, and any polynomial obtained by multiplying $\Phi_5(z) \Phi_6(z)$ by a product of such polynomials will evaluate to 0, 2, 3, and 5 at $\zeta$, $\zeta^2$, $\zeta^3$, and 1, respectively. But none of the combinations I've tried gives rise to a polynomial with nonnegative coefficients.