Looping through zeroes in polynomials Question.

If $p$ is a polynomial of degree $n$ with $p(\alpha)=0$, what do we know of the polynomial $q$ (with degree $n-1$) such that the numbers $(q^k(\alpha))_{k=1}^n$ contain all of the zeroes of $p$?

Here I denote $q(q(\cdots q(\alpha)))=q^k(\alpha)$.

Notes.
We know for a fact such $q$ exists, since there always exists a polynomial of degree $n-1$ through $n$ given points. $q$ is not unique, however, since there are multiple permutations we can put the zeroes in.

Examples.
For linear $p$ (write $p(x)=a_1x+a_0$), this is obvious; $q(x)=\tfrac{-a_0}{a_1}=\alpha$ suffices. If $p(\alpha)=0$, then $q(\alpha)=\alpha$ indeed are all the zeroes of $p$.
If $p$ is quadratic, write $p(x)=a_2x^2+a_1x+a_0$, and have $p(\alpha)=0$ again; now $q(x)=\frac{-a_1}{a_2}-x$.
If $p$ is cubic, write $p(x)=a_3x^3+a_2x^2+a_1x+a_0$. This is where I get stuck, since the roots of cubic equations aren't expressions that are easy to work with.

Attempts.
First I see (denote the (not necessarily real) zeroes by $z_1,z_2,\cdots,z_n$) that $z_1+\cdots+z_n=\frac{-a_{n-1}}{a_n}$ and $z_1z_2\cdots z_n=\frac{(-1)^na_0}{a_n}$. We can produce similar expressions for the other coefficients, but I doubt this is useful; they're not even solvable for $n>4$. We also have (given $z_1$)
$$-a_nz_1^n=a_{n-1}z_1^{n-1}+\cdots+a_1x+a_0$$
with which we can reduce every expression of degree $n$ or larger in $z_1$ to an expression of degree $n-1$ or smaller.
For $n=3$ (let's do some specific examples), we could write $q(x)=b_2x^2+b_1x+b_0$, and take for example $p(x)=x^3-x-1$. Then, if $\alpha$ is a zero of $p$, then $\alpha^3=\alpha+1$, and so $q(\alpha)^3=q(\alpha)+1$, which is
$$(b_2\alpha^2+b_1\alpha+b_0)^3=b_2\alpha^2+b_1\alpha+b_0+1$$
working out the constant terms gives $b_2^3+b_1^3+b_0^3+6b_0b_1b_2=b_0+1$ which isn't very useful either.
Please, enlighten me. Has there been done work on this subject, am I missing something obvious, or perhaps you see something that I missed?

Let $\alpha=\alpha_1, \alpha_2, \ldots, \alpha_m$ be the distinct roots of $p$.
Choose a permutation $\sigma$ of $1,2,\dots,m$ without fixed points. For instance, an $m$-cycle such as $(12\cdots m)$.
Let $q$ be the unique polynomial such that $q(\alpha_i)=\alpha_{\sigma(i)}$. That will work, but won't have degree $n-1$.
