Let $\phi(x,y,z)$ be an alternating trilinear form on a space $V$ over a field $K$.

Let $u \in \mathbb{P}(V)$ be a projective point over $V$, then we say that the rank of $u$ is equal to the rank of the alternating bilinear form $\phi(u,.,.)$.

To the form $\phi$ we can associate a graph $G_\phi$, whose the vertices are the projective points in $\mathbb{P}(V)$, and two vertices $u,v$ share an edge if the linear form $\phi(u,v,.)$ is identically zero.

For Section 7 of my work, I computed the restriction of these graphs to points of rank 4 for uniformly random $9$-dimensional alternating trilinear forms over finite fields of prime order $p$, and I stumbled on the following observation, which I cannot explain:

It seems that with high probability (the probability seems to go to 1 as $p$ grows), the graphs have Dihedral symmetry $D_{N}$, where $N$ is the number of points of rank 4. See, e.g., pictures below over GF(5).

Where does this come form? A guess would be that the automorphisms of the graph come from automorphisms of the trilinear form, but this is not the case. Usually, random forms $\phi$ have only a few (e.g. 2,3,4, or 6) automorphisms, but the graphs have much more automorphisms.

In fact, the automorphism group of the graph is not (always) a subgroup of $GL(n,K)$, because the rotation of the second picture below has order 29, but 29 does not divide $|GL(9,5)| = 2^{25}×3^5×5^{36}×7×11×13^2×19×31^3×71×313×829×19531$.

It is not so difficult to prove that the average (for random $\phi$) number of nodes in the graph is $q^2 + O(q)$, and the average number of edges is $q^3/2 + O(q^2)$ (Theorem 1 in my work). But the structure of the graph is still very mysterious to me. The pictures say that there is a free and transitive group action of the cyclic group of order $N$ on the $N$ points of rank $4$, but I have no idea what this group is and how it acts.

Graph-theorethic Algorithms for the Alternating Trilinear Form Equivalence problem$\endgroup$