I asked this question at MSE but I did not receive an answer. So I ask it at MO:

We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:

Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$

I asked this question at MSE but I did not receive an answer. So I ask it at MO:

We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:

Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$

I asked this question at MSE but I did not receive an answer. So I ask it at MO:

We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:

Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$

I asked this question at MSE but I did not receive an answer. So I ask it at MO:

We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:

Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$

I asked this question at MSE but I did not receive an answer. So I ask it at MO:

We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:

Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$

I asked this question at MSE but I did not receive an answer. So I ask it at MO:

We denote the field of rational numbers by $\mathbb{Q}$. The Galois group of a polynomial $f$ is denoted by $Gal(f)$. The commutator subgroup of a group $G$ is denoted by $G'$.

Is there a $\mathbb{Q}$-linear map $T$ over $\mathbb{Q}[x]$ such that for all polynomials $f\in \mathbb{Q}[x]$ we have $$Gal(T(f))=(Gal(f))'$$

For a related post see the following question:

Endofunctors on the category of groups which are Galois- related to a linear map on $\mathbb{Q}[x]$