Let $F(x,y) = a_6 x^6 + a_5 x^5 y + \cdots + a_0 y^6 \in \mathbb{Z}[x,y]$ be a binary sextic form with non-zero discriminant. By applying a $\operatorname{GL}_2(\mathbb{Z})$-translation if necessary, we can assume that $a_0 \ne 0$. Then $F$ is said to be a *Klein form* if its coefficients satisfy the quadratic equations

$$\displaystyle 10 a_0 a_4 - 5 a_1 a_3 + 2a_2^2 = 0,$$ $$\displaystyle 25 a_0 a_5 - 5 a_1 a_4 + a_2 a_3 = 0,$$ and $$\displaystyle 50 a_0 a_6 - 2 a_2 a_4 + a_3^2 = 0.$$

My question is, what is the Galois group of the splitting field of a generic binary sextic Klein form?

It is easy to show that the generic Galois group cannot be the full symmetric group $S_6$, by a theorem of Bhargava and Yang (see Theorem 5 in https://arxiv.org/abs/1312.7339), since Klein forms have non-trivial automorphisms in $\operatorname{PGL}_2(\overline{\mathbb{Q}})$. In fact, they have the largest possible $\operatorname{PGL}_2(\overline{\mathbb{Q}})$-automorphism group of any sextic form with non-zero discriminant; so one would expect that their Galois group is very small. In particular, its Galois group should at least be solvable.