Let $\mathbb{A}^3 = \mathbb{F}\_q^3$. Consider the following three functions $\mathbb{A}^3\to\mathbb{A}^3$:
\begin{eqnarray\*}
h: (x, y, z) &\mapsto& (x, y, xy - z) \\
u: (x, y, z) &\mapsto& (y, x, z) \\
v: (x, y, z) &\mapsto& (x, z, y)
\end{eqnarray\*}
It immediately follows that those three morphisms are involutions, and hence are permutations in $Sym(\mathbb{A}^3) \cong S\_{q^3}$.


If we have enough points in the space, it follows that $\langle h, u \rangle \cong \mathbb{Z}\_2 \times \mathbb{Z}\_2$ (as $h$, and $u$ commute), $\langle u, v\rangle \cong S\_3$ (we can identify $u$ and $v$ with the permutations $(12)$ and $(23)$), and $\langle h, v\rangle \cong D\_4$ (exhaustion). How can I figure out $\langle h, u, v \rangle$? It is some finite quotient of $\mathbb{Z}\_2 \ast \mathbb{Z}\_2 \ast \mathbb{Z}\_2$.


Denote the Fricke polynomial by $$\kappa(x, y, z) = x^2 + y^2 + z^2 - xyz - 2.$$ It is surjective (since any element in a finite field can be written as a sum of two squares). So the solutions $\kappa^{-1}(a)$ partition the affine space. The permutations $h, u, v$ fix each level set $\kappa^{-1}(a)$, so the action is not transitive, orbits stay within the same level sets.


It interesting to note that $Aut(\kappa)$ is generated by $h, u, v$ and maps that change the signs of pairs of coordinates.


The structure of the generated permutation group will depend on $q$. Based on computational experiments, it appears that the permutation $(x, y, z) \mapsto (z, y, yz -x)$ will have order $p(p^{2n}-1)/2$ over $\mathbb{F}\_{p^n}$ for $p$ odd, and $2(2^{2n}-1)$ for $p = 2$.



What is the permutation group generated by those three given morphisms of the affine space $\mathbb{F}_q^3$?
The experimental observation about the order of $w:(x, y, z) \mapsto (z, y, yz-x)$ isn't hard to verify: By induction, one shows that $w^m(x,y,z)=(a\_mx+b\_mz,y,c\_mx+d\_mz)$ with $\begin{pmatrix}a\_m&b\_m\\c\_m&d\_m\end{pmatrix}=\begin{pmatrix}0&1\\-1&y\end{pmatrix}^m$. Looking at the eigenvalues of $\begin{pmatrix}0&1\\-1&y\end{pmatrix}$ (which are in $\mathbb F\_{q^2}$), and not forgetting the non-diagonalizable cases $y=2$ and $y=-2$, yields the assertion after a short calculation.


