Show that the map on $2\times 2$ matrices
\begin{eqnarray} \left( \begin{matrix} a & b\\ c & d \end{matrix} \right)\overset{\Phi}{\mapsto} \left( \begin{matrix} a & b\\ c & d \end{matrix} \right)\cdot \left( \begin{matrix} 2 (a c - b^2) & a d - b c\\ a d - b c & 2(b d - c^2) \end{matrix} \right)^{-1} \cdot \left( \begin{matrix} 0 & 1\\ 1 & 0 \end{matrix} \right) \end{eqnarray}
is an involution, that is $\Phi\circ \Phi=\operatorname{Id}$
Notes:
For convenience, one could treat $a$, $b$, $c$, $d$ as independent variables, with coefficients in $\mathbb{C}$.
If we consider the form $F(x,y) =a x^3 + 3 b x^2 y + 3 c x y^2 + d$, then the Hessian determinant is a quadratic form with matrix coefficients (up to a constant( $\left( \begin{matrix} 2 (a c - b^2) & a d - b c\\ a d - b c & 2(b d - c^2) \end{matrix} \right)$. Its determinant is up to a constant the discriminant of the cubic form $F$. See also the matrix $\left(\begin{matrix} a& b &c \\ b&c&d\end{matrix}\right)$ and its $2\times 2$ minors.
If instead of the inverse on the second factor we consider the adjugate, then we don't have to worry about denominators. Then the composition will be a constant times the matrix.
One could consider an alternate map with the third factor on RHS given by $\left( \begin{matrix} 0 & -1\\ 1 & 0 \end{matrix} \right)$ . This is connected to the covariant $t$ for cubics, and an involution on triplets ( the roots of the cubic). Note that the square of the map is in this case $-\operatorname{Id}$.
One can show that we have an involution by direct calculation ( RHS could be multiplied by an arbitrary fixed constant). It would be nice to have a proof involving some shortcuts. Maybe everything is just something basic.
Any feedback would be appreciated!
