Let $f(x)$ be a polynomial with integer coefficients and non-zero discriminant (so each root of $f$ has multiplicity one). Consider the projective roots $\theta_1, \cdots, \theta_n$ of $f$, as elements of the projective line $\mathbb{P}^1(\mathbb{C})$. We say that the roots of $f$ have symmetry if there exists an element in $T \in \operatorname{GL}_2(\mathbb{C})$, not proportional to the identity, such that $T$ fixes the set of roots (i.e., permutes the roots) via Mobius action. Observe that the roots of polynomials of degrees 2,3,4 always have symmetry, but generic configurations of five or more points in $\mathbb{P}^1$ typically have no symmetry.
It is quite clear that symmetry and the Galois group $\operatorname{Gal}(f)$ are related. In fact Bhargava and Yang proved the following theorem:
Let $f$ be a polynomial of degree $n \geq 5$ and integer coefficients. If the roots of $f$ have symmetry over $\operatorname{PGL}_2(\overline{\mathbb{Q}})$, then the Galois group of $f$ cannot be isomorphic to the symmetric group $S_n$.
Are there results in the literature which studies the relationship between symmetry of roots and the Galois group? Presumably, the more symmetric the roots (i.e., the more non-trivial automorphisms) the smaller the Galois group is forced to be.
