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?