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:

 1. $f(ax) = af(x)$ for all $a \in GF(q)^\times$,
 2. $Tr_{K/F}(x f(x)) = 0$. ($F = GF(q)$)

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 $Aut(PG(2,q))$ on $GF(q^3)$ [thinking of GF(q^3) as a 3-dimensional vector space over $GF(q)$] and call two permutation polynomials isomorphic if they belong to the same orbit of this action. 

[1] http://homepages.cwi.nl/~lex/files/countpms2.pdf