Classification conjecture for quadratic permutation systems over finite fields of odd characteristic

The conjecture is false for EVERY odd prime power $q$ and EVERY $n\ge3$, under precisely the polynomial-system equivalences defined in the source.

Let $K=\mathbb F_q$, and choose a nonsquare $\delta\in K^\times$. Define

$$F(x,y,z,x_4,\ldots,x_n) =\bigl(x+zy,\ y+\delta zx,\ z,\ x_4,\ldots,x_n\bigr).$$

For each fixed $z\in K$, the transformation on $(x,y)$ has matrix

$$B_z= \begin{pmatrix} 1&z\\ \delta z&1 \end{pmatrix}, \qquad \det B_z=1-\delta z^2.$$

Since $\delta$ is a nonsquare, $1-\delta z^2\ne0$ for every $z\in K$. Thus each $z$-fiber is mapped bijectively to itself, and the remaining coordinates are fixed. Hence $F$ permutes $K^n$.

Its FORMAL Jacobian determinant, however, is

$$\det J_F=1-\delta z^2,$$

which is a nonconstant polynomial.

Having a nonzero constant formal Jacobian determinant is invariant under both equivalences in Definitions 3.1 and 3.2. Indeed, if

$$G=\rho\circ F\circ\sigma$$

for nonsingular linear transformations $\rho,\sigma$, the chain rule gives

$$\det J_G(\mathbf x) =(\det\rho)(\det\sigma)\det J_F(\sigma\mathbf x).$$

An invertible linear substitution preserves whether this polynomial is a nonzero constant.

For a coordinate-shift equivalence, the two systems have the source's exact forms

$$F_0=(f_1(x_1,\ldots,x_k),\ldots,f_k(x_1,\ldots,x_k), x_{k+1},\ldots,x_n)$$

and

$$G_0=(f_1(x_1,\ldots,x_k),\ldots,f_k(x_1,\ldots,x_k), x_{k+1}+h_{k+1}(x_1,\ldots,x_k),\ldots, x_n+h_n(x_1,\ldots,x_{n-1})).$$

Their Jacobian matrices are block lower triangular with the SAME upper-left $k\times k$ block. The lower-right block is respectively the identity or a unit lower-triangular matrix. Therefore

$$\det J_{G_0}=\det J_{F_0}$$

as formal polynomials. Coordinate permutations preserve the same constant-determinant property.

The identity system has Jacobian determinant one, so EVERY polynomial system equivalent to it under the specified operations has nonzero constant Jacobian determinant. Our permutation $F$ has nonconstant determinant $1-\delta z^2$, and therefore cannot be equivalent to the identity.

Already over the smallest admissible field, take $q=3$, $\delta=2=-1$, and $n=3$:

$$F(x,y,z)=(x+zy,\ y-zx,\ z), \qquad \det J_F=1+z^2.$$

For $z=0,1,2$, the fiber determinants are $1,2,2$. Hence $F$ permutes all 27 points of $\mathbb F_3^3$ but is not equivalent to the identity. This refutes the conjecture at its smallest permitted field and dimension.