My question: Let $f(x)$ be a polynomial with Gaussian integer coefficients. Under what conditions does $f(x^n) \mid (f(x))^n$? Can we generate all such polynomials given $d$, the degree of $f(x)$?
My work: I can't think of a good way to approach this just yet and have an essay to write. This trivially holds for monomials with integer coefficients, and seems to hold for all monomials with coefficients in $\mathbb{Z}[i]$. Beyond that, I suspect this question can be attacked by inspecting its roots, and has something to do with the cyclotomic polynomials. But that's just my intuition; this is no problem I found in a textbook and may be unapproachable. However, while fiddling around with this problem, I did come up with some extension questions.
Extension #1: Are there infinitely many non-monomial $f(x)$ for every $n \in \mathbb{N}$? Given that I struggle to find any such $f(x)$, I would say no (in fact, it may be the case that there are no such non-monomial $f(x)$). Perhaps there are only infinitely many $f(x)$ for certain $n$.
Extension #2: A similar, perhaps easier (but not easy) question: when does $f(x)$ divide $f(x^n)$? Let $d$ be the degree of $f(x)$. For any choice of $d$ and $n$, are there always infinitely many $f(x)$ satisfying $f(x) \mid f(x^n)$? The answer is almost definitely no. Letting $n,d=2$ we can write out pretty quickly all such $f(x)$, and I am sure a proof would show there exist no more.
Extension #3: When does $f(x^k) \mid (f(x))^n$ for $0<k<n$? Again, can we generate all such polynomials given $n$ and the degree, $d$, of $f(x)$?
If $f(x)|f(x^n)$, it means for any root r of f(x), we have $f(r^n)=0$ too so that r^n is root is f(x). So it means that |r|=0 or |r|=1, otherwise we will be able to generate infinity number of roots. So f(x) must be multiplication of x^k and cyclotomic polynomial