Find all the monic polynomials P(x) ∈ C [X] satisfying the below condition What are all the monic polynomials $P(X) \in \mathbb C [X]$, with simple roots such that $$P(X^n) = \pm P(X) P(\zeta X) P (\zeta^2 X) …P (\zeta^{n-1} X),$$ where ζ is a primitive n-th root of unity?
 A: This grew to long for a comment so I post it as an answer.

Let $P$ be any monic polynomial with only simple roots $r_1,\dots, r_n$. Then, $$P(X)=\prod\limits_{i=1}^n (X-r_i).$$ Now, $$P(X^n)=\prod\limits_{i=1}^n (X^n-r_i).$$ Thus, $$\prod\limits_{i=1}^{n-1} P(\zeta^i X)=\prod\limits_{i=0}^{n-1} \prod\limits_{j=1}^n(\zeta^iX-r_j)=\prod\limits_{j=1}^n(X^n-r_i)=P(X^n).$$ The roots of $P(X^n)$ are the $n$-th roots of the $r_i$'s. So what you are asking for is an equality of products $$\prod\limits_{i=0}^{n-1} \prod\limits_{j=1}^n(\zeta^iX-r_j)=\prod\limits_{i=1}^n\prod\limits_{j=0}^{n-1}(X-\zeta^j\sqrt[n]{r_i}).$$ Now compare the factors. On the RHS, the roots are the $n$-th roots of the roots of $P$. On the LHS the roots are $r_j\zeta^{-i}$. So in order to have equality, the roots need to coincide and you see that this is not necessarily true. In particular, you will want to allow now monic polynomials by polynomials with some root of unity as factor (to bring the LHS to a nicer form) and it also seems that $\sqrt[n]{r_i}=r_j$ should be a condition.
