Suppose we have a polynomial $z \to f(z,c)$ defined over $\mathbb Z$ with a free parameter $c$, for instance $z \to z^2 + c$ and we consider the iterates $z \to f^{(n)}(z,c)$ and define the polynomials $g_n(c) = f^{(n)}(0,c)$. That is, the roots of $g_n$ correspond to those $c$ for which $0$ has a period of size dividing $n$. For example, with $f(z) = z^2 + c$: \begin{gather*} g_2(c) = c^2 + c \\ g_3(c) = (c^2 + c)^2 + c \end{gather*} and so on. We also define $h_n(x) \mid g_n(x)$ to be the polynomial with roots corresponding to $c$ so that $0$ has period exactly $n$. Then: \begin{gather*} h_2(c) = c+1 \\ h_3(c) = c^3 + 2c^2 + c + 1 \end{gather*} and so on.

In particular, if $f(z)$ has degree $d$, then $g_n(z)$ has degree $d^{n-1}$ and by möbius inversion: $$\deg(h_n) = \sum_{m\mid n}\mu\left(\frac{n}{m}\right)d^{n-1}.$$

Now it turns out to be true that we have the following congruence: $$\sum_{m\mid n}\mu\left(\frac{n}{m}\right)d^n \equiv 0 \pmod{n}$$ and therefore, at least when $\gcd(d,n) = 1$, we have that $\deg(h_n)$ is divisible by $n$. Is there a natural way to partition the roots of $h_n$ (corresponding to $c$ so that $0$ has an orbit of period exactly $n$) into sets of size $n$ (or perhaps $n$ sets?)?

I don't see a straightforward way of doing this, especially because $h_3(c) = c^3 + 2c^2 + c + 1$ in the example above turns out not to generate a Galois extension and so we cannot write all 3 roots as algebraic expressions in one of the roots.

(I believe it is a conjecture that the $h_n$ are irreducible over $\mathbb Q$ at least for $f(z) = z^2 + c$. I would be very happy if someone could provide a reference for this conjecture and also perhaps what we expect to happen in general.)