https://arxiv.org/pdf/1211.6044.pdf Theorem 1.14 (2): The monomial $x^n$ is a permutation polynomial of $F_q$ if and only if $gcd(n, q − 1) = 1$.
Let's pick $n=2$ and $q=6$ so $gcd(2,5) = 1$ in this case $x^2$ should be a permutation polynomial over $F_6$. Let's put them into a table.
| 0 | 1 | 2 | 3 | 4 | 5 | |
|---|---|---|---|---|---|---|
| $x^2$ | 0 | 1 | 4 | 3 | 4 | 1 |
As we see the function is not even one-to-one, it is not a permutation polynomial, what am I missing here?
[Update] Thanks to @GerryMyerson, my example definition of field is wrong here. Modulo over 6 have zero divisor. If I try it with a $F_p$ over modulo p and p is prime then it works.
$F_7$ which is modulo over 7 and $x^5$ should be permutation polynomial because of $gcd(5,6)=1$.
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|---|
| $x^5$ | 0 | 1 | 4 | 5 | 2 | 3 | 6 |