I thank J.C. and Arturo Magidin for interesting suggestions in the comments to this question

In this post the field of rational numbers is denoted by $\mathbb{Q}$. The space of polynomials with rational coefficients is denoted by $\mathbb{Q}[x]$. The commutator subgroup of a group $G$ is denoted by $G'$. The isomorphism between two groups is denoted by $\simeq$

We learn from the answer to the following question

Is there a $\mathbb{Q}$-linear map $T$ on $\mathbb{Q}[x]$ such that for all polynomials $\mathrm{Gal}(T(f))$ is the commutator subgroup of $\mathrm{Gal}(f)$?

that there does not exist a $\mathbb{Q}$-linear map $T$ on $\mathbb{Q}[x]$ such that for all $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))\simeq(Gal(f))'$$

Motivated by the above fact we ask the following question:

Assume that $T$ is a a linear map over $\mathbb{Q}[x]$, and $\mathcal{F}$ is an endofunctor on the category of groups. we say that $T$ is Galois-related to $\mathcal{F}$ if we have $$Gal(T(f))\simeq\mathcal{F}(Gal(f))\;\;\;\text{For all polynomial } f \in \mathbb{Q}[x]$$

What, if any, are some non trivial examples of such situation? (A non identity functor?Some interesting examples?)