Prove there exists at most 2 quadratics permuting a set of complex numbers Let $S=\{s_1,s_2,\dots,s_n\}$ such that $s_i\in\mathbb{C}$ for all $1\leq i\leq n$. Given that $n\geq 9$ and exactly $n-3$ of the $s_i$ are real numbers, prove that there are at most two quadratic polynomials $f(z)$ with complex coefficiants such that $f(S)=S$, i.e. $f$ permutes the elements of $S$.
This question seems to have a very strange condition, namely that $n\geq 9$ and exactly $n-3$ of the $s_i$ are real. However, what I tried was: without loss of generality let $f(s_i)=s_{i+1}$ for $1\leq i \leq n$ (here $s_{n+1}=s_1$). Also, let $f(z)=a_2z^2+a_1z+a_0$. So we have
$$\begin{cases} a_2s_1^2+a_1s_1+a_0=s_2 \\ a_2s_2^2+a_1s_2+a_0=s_3 \\ a_2s_2^2+a_1s_2+a_0=s_3 \\ \vdots\\a_2s_n^2+a_1s_n+a_0=s_1 \end{cases}$$
But $s_1,s_2,\dots s_n$ are constants, so this is just a system of $n$ linear equations in $a_2,a_1,a_0$?? So we just have to consider cases where some of the equations are linear multiples of each other, but this can't be the case due to $3$ of the $s_i$ being complex but not real.
I'm not sure if this argument works at all; note that I did not use $n\geq 9$. I would appreciate it if someone could correct my possible mistakes and give a complete solution. 
 A: If $f$ is a quadratic polynomial over $\mathbb C$, then the imaginary part of $f(x)$ for real $x$ is a polynomial function in $x$ of degree $\le 2$. That means that either $f(x)$ is real for all real $x$ or there are at most two real $x$ such that $f(x)$ is real.
In the latter case most $2$ of the $n-3$ real points in $S$ can be images of other real points, so there are at most $5$ real points in $f(S)$ -- which means that $f$ doesn't permute $S$, because $S$ itself has $n-3\ge 6$ real points.
Therefore $f$ has to map every real argument to a real value, which again means that $f$ has real coefficients.
Therefore neither of the non-real points in $S$ can be images of any of the real points, so $f$ permutes the three points in $S\setminus \mathbb R$.
There are then several different cases:


*

*$f$ fixes all three non-real points. In that case $f$ has to be the identity, which I assume does not count as a quadratic.

*$f$ fixes one of the non-real points, say $s_1$, and interchanges the two others, say, $s_2$ and $s_3$. In that case all three points are solutions of $f(f(z))=z$, which is a fourth-degree equation with real coefficients. But there are at most 4 solutions to that, which come in conjugate pairs, so $s_2$ and $s_3$ are conjugates. But in that case $f$ must be the only quadratic permutation of $S$, because an $f$ with real coefficients that maps either $s_2$ or $s_3$ to $s_1$ must map its conjugate to $\bar s_1$ which is not in $S$.

*Otherwise $f$ permutes $\{s_1,s_2,s_3\}$ cyclically. There are exactly two ways to do this, and each of them fixes all coefficients on $f$.

Note that this argument doesn't show that there is any $S$ satisfying the conditions that is actually permuted by two different quadratics, and it is not clear to me that there is. It is easy enough to see that there must be many examples where there is at least one quadratic that permutes $S$ and permutes the three non-reals cyclically, but I'm not even sure whether in any of them the reverse cycle will even have real coefficients!
