Given a prime power $q$, I would like to enumerate (preferably up to isomorphism*) all the permutation polynomials $f(x)$ on $K = GF(q^3)$ satisfying the following conditions:
- $f(ax) = af(x)$ for all $a \in GF(q)^\times$ and $f(0) = 0$,
- $Tr_{K/F}(x f(x)) = 0$. ($F = GF(q)$)
I know that for for $q = 3$ there are $5$ isomorphism classes and in total there are $3852$ such polynomials. For $q = 4$, there are $18534400$ of them. This is from computer computations.
It is not hard to see that such polynomials correspond to perfect matchings in the incidence graph of the projective plane $PG(2,q)$. By [1] a $k$-regular bipartite graph of size $2n$ has at least $$\left( \frac{(k-1)^{(k-1)}}{k^{(k-2)}} \right)^n$$ perfect matchings. Therefore there is a lower bound of $\frac{q^{qn}}{(q+1)^{n(q-1)}}$ where $n = 1 + q + q^2$.
To the finite field experts: is there a way of theoretically classifying such polynomials? Can we give some bounds on the total number? Any suggestions on what approach might work?
*Look at the action of $\Gamma L(3,q)$ on $GF(q^3)$ viewed as a vector space over $GF(q)$. For a permutation polynomial $f$ and a group element $\sigma$ define $f^\sigma (x) = \hat{\sigma} (f(\sigma(x))$ where $\hat{\sigma}$ is the adjoint of $f$ w.r.t. the symmetric bilinear form $(x, y) \mapsto Tr_{K/F}(xy)$.