Let $f = f_2 x^2 + f_1 xy + f_0 y^2$ and $g = g_2 x^2 + g_1 xy + g_0 y^2$ be two binary quadratic forms with co-prime integer coefficients and non-zero discriminants. Suppose further that $\Delta(f,g) = 2 f_2 g_0 - f_1 g_1 + 2 f_0 g_2 = 0$. Then the matrices

$\displaystyle M_f = \begin{pmatrix} f_1 & 2 f_0 \\ - 2 f_2 & - f_1 \end{pmatrix}, M_g = \begin{pmatrix} g_1 & 2 g_0 \\ -2 g_2 & - g_1 \end{pmatrix}$

generate a subgroup isomorphic to $C_2 \times C_2$ in $\operatorname{PGL}_2(\mathbb{R})$. By scaling appropriately, they also generate a subgroup isomorphic to the dihedral group $D_4$ of order 8 in $\operatorname{GL}_2(\mathbb{R})$.

It is well-known that finite subgroups of $\operatorname{GL}_2(\mathbb{R})$ are conjugate to finite subgroups of $\operatorname{O}_2(\mathbb{R})$; indeed, all finite groups that are isomorphic as groups are conjugate, with the exception of groups of order 2 (the group $\{\pm I_{2 \times 2}\}$ is conjugate only to itself, and all other copies of $C_2$ in $\operatorname{GL}_2(\mathbb{R})$ are conjugate to each other). In particular, the group $G_{f,g}$ generated by $M_f, M_g$ (appropriately scaled to have determinant $\pm 1$) is $\operatorname{GL}_2(\mathbb{R})$-conjugate to all other groups in $\operatorname{GL}_2(\mathbb{R})$ isomorphic to the dihedral group $D_4$.

My question concerns a more restricted notion of conjugacy. Let $(u,v)$ be another pair of binary quadratic forms with co-prime integer coefficients and such that $\Delta(f) = \Delta(u)$, $\Delta(g) = \Delta(v)$, $\Delta(u,v) = \Delta(f,g) = 0$. Then does there exist a matrix $T$ of the shape

$T = \frac{1}{\sqrt{|t_1 t_4 - t_2 t_3|}} \begin{pmatrix} t_1 & t_2 \\ t_3 & t_4 \end{pmatrix}, t_1, t_2, t_3, t_4 \in \mathbb{Z}, t_1 t_4 - t_2 t_3 \ne 0$

such that $G_{f,g}, G_{u,v}$ are conjugate by $T$?

An equivalent way to formulate this question is as follows: let $(f,g), (u,v)$ be two pairs of primitive integral binary quadratic forms with non-zero discriminant, such that the joint discriminant $\Delta(f,g) = \Delta(u,v) = 0$ and $\Delta(f) = \Delta(u), \Delta(g) = \Delta(v)$. Then does there exist a matrix $T$ as above such that $f_T = \pm u, g_T = \pm v$?, where the subscript denotes substitution by $T$?