Values of a pair of determinants Let $\mathbf{x} = (x_0, x_1, x_2), \mathbf{y} = (y_0, y_1, y_2)$ be vectors over a field $\mathbb{F}$ of characteristic zero. Define the function
$\displaystyle S(\mathbf{x}, \mathbf{y}) = x_2 (y_0^2 - 2 y_1 y_2) + x_1 (2 y_2^2 - y_0 y_1) + x_0 (y_1^2 - y_0 y_2) = \begin{vmatrix} x_2 & x_1 & x_0 \\ y_2 & y_1 & y_0 \\ y_1 & y_0 & 2 y_2 \end{vmatrix}$
and $T(\mathbf{x}, \mathbf{y}) = S(\mathbf{y}, \mathbf{x})$.
Curiously, I found that for fixed $(s,t) \in \mathbb{F}^2$ the set of solutions to $s = S(\mathbf{x}, \mathbf{y}), t = T(\mathbf{x}, \mathbf{y})$ is stable under the map
$\mathbf{x} \mapsto \begin{bmatrix} x_0 + 2 x_1 + 2 x_2 \\ x_0 + x_1 + 2 x_2 \\ x_0 + x_1 + x_2 \end{bmatrix}, \mathbf{y} \mapsto \begin{bmatrix} y_0 + 2 y_1 + 2 y_2 \\ y_0 + y_1 + 2 y_2 \\ y_0 + y_1 + y_2 \end{bmatrix}.$
Moreover, the matrix defining this linear map which is
$M = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \\ 1 & 1 & 1 \end{bmatrix}$
has determinant one.
Thus, if we define the group $\mathcal{G} \subset \text{GL}_3(\mathbb{F})$ to be the set of $3 \times 3$ matrices $A$ over $\mathbb{F}$ such that $S(\mathbf{x}, \mathbf{y}) = S(A \mathbf{x}, A \mathbf{y})$ for all $\mathbf{x}, \mathbf{y} \in \mathbb{F}^3$, then we have shown that $M \in \mathcal{G}$. Further, $M$ has infinite order so $\mathcal{G}$ contains infinitely many elements.
Is it possible to determine $\mathcal{G}$ in a reasonable way?
 A: First, by transposing and interchanging two rows $S({\bf x},{\bf y}) = -\left|\begin{matrix} x_0 & y_0 & 2y_2 \\ x_1 & y_1 & y_0 \\ x_2 & y_2 & y_1  \end{matrix} \right|$. Notice also that if $T = \left[\begin{matrix}0&0&2 \\ 1&0 &0 \\ 0& 1& 0\end{matrix} \right]$ then $T{\bf y} = \left[\begin{matrix} 2y_2 \\ y_0 \\ y_1\end{matrix}\right]$. Hence, $S({\bf x},{\bf y}) = -\left| {\bf x}\ \ {\bf y}\ \ T{\bf y} \right|$
Suppose $A\in \mathcal G$ then for all ${\bf x},{\bf y} \in \mathbb F^3$
$$
S({\bf x},{\bf y}) = S(A{\bf x}, A{\bf y}) = -\left|{\bf x}\ \ {\bf y} \ \ |A|A^{-1}TA{\bf y}\right|.
$$
Now when ${\bf x} = e_3$ then this gives
$$
y_0^2 - 2y_1y_2 = y_0\langle|A|A^{-1}TA{\bf y}, e_2\rangle - y_1\langle|A|A^{-1}TA{\bf y}, e_1\rangle.
$$
Subsequently, when ${\bf y} = e_1$ this gives $1 = \langle|A|A^{-1}TAe_1, e_2\rangle$, and when ${\bf y} = e_2$ this gives $0 = \langle|A|A^{-1}Tae_2, e_1\rangle$. Moreover, when ${\bf y} = e_1+e_3$ then after some computation $$\langle|A|A^{-1}TAe_1, e_1\rangle = \langle|A|A^{-1}TAe_3, e_3\rangle$$
When ${\bf x} = e_2$ this gives
$$
2y_2^2 - y_0y_1 = y_2\langle|A|A^{-1}TA{\bf y}, e_1\rangle - y_0\langle|A|A^{-1}TA{\bf y}, e_3\rangle,
$$
${\bf y} = e_1$ gives $0 = \langle|A|A^{-1}TAe_1, e_3\rangle$, and ${\bf y} = e_3$ gives $2 = \langle|A|A^{-1}Tae_3, e_1\rangle$.
Moreover, when ${\bf y} = e_1+e_2$ then $$\langle|A|A^{-1}TAe_1, e_1\rangle = \langle|A|A^{-1}TAe_2, e_2\rangle$$
When ${\bf x} = e_1$ this gives
$$
y_1^2 - y_0y_2 = y_1\langle|A|A^{-1}TA{\bf y}, e_3\rangle - y_2\langle|A|A^{-1}TA{\bf y}, e_2\rangle
$$
${\bf y}=e_2$ gives $1 = \langle|A|A^{-1}TAe_2, e_3\rangle$ and ${\bf y} = e_3$ gives $0 = \langle|A|A^{-1}TAe_3, e_2\rangle$
If we define $\lambda := \langle|A|A^{-1}TAe_1, e_1\rangle$ then the above argument concludes
$$
|A|A^{-1}TA = \left[\begin{matrix}\lambda&0&2 \\ 1&\lambda &0 \\ 0& 1& \lambda\end{matrix} \right] = \lambda I + T
$$
and so
$$
TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT.
$$
One won't get anymore out of these equations. Furthermore, any $A$ satisfying this equation gives
$$
S(A{\bf x}, A{\bf y}) = -|{\bf x} \ \ {\bf y} \ \ \lambda{\bf y} + T{\bf y}| = S({\bf x}, {\bf y})
$$
by column replacement. Your $M$ corresponds to the case where $\lambda = 0$, that is commutes with $T$, and $|A| = 1$. However, there may be more matrices in $\mathcal G$.
Therefore,
$$
\mathcal G = \left\{ A\in {\rm GL}_3(\mathbb F) : TA = \frac{1}{|A|}\lambda A + \frac{1}{|A|}AT, \lambda\in \mathbb F  \right\}
$$
