**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 

 "http://mathoverflow.net/questions/219076/a-question-in-galois-theory " that:



**There is  no  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  are some  non trivial examples of such situation? (A  non identity  functor?Some interesting examples?)