The answer is no. First we see that $T$ is injective: Let $0\ne g\in\mathbb Q[X]$ with $T(g)=0$. Then $\textrm{Gal}(f-\alpha g)'=\textrm{Gal}(f)'$ for all $f\in\mathbb Q[X]$ and $\alpha\in\mathbb Q$. Pick $\beta\in\mathbb Q$ with $g(\beta)\ne0$, and $f\in\mathbb Q[X]$ of degree $n$ larger than $\deg g$ and $2$ such that $\textrm{Gal}(f)=S_n$. Then $\textrm{Gal}(f)=S_n'=A_n$. But the polynomial $f(X)-\frac{f(\beta)}{g(\beta)}g(X)$ has the root $\beta$, so its Galois group is a subgroup of $S_{n-1}$. But no subgroup of $S_{n-1}$ is isomorphic to $A_n$.

Thus we find $f\in\mathbb Q[X]$ of degree $2$ such that $F=T(f)$ has degree at least $2$. Since $\textrm{Gal}(F)=\textrm{Gal}(f)'=1$, $F$ splits into linear factors. Next pick $\gamma\in\mathbb Q$ such that $T(X+\gamma)=T(X)+\gamma T(1)$ has no common root with $F$. Then $F$ and $G=T(X+\gamma)$ are relatively prime. As $f+\alpha(X+\gamma)$ has degree at most $2$ for all rational $\alpha$, we see that $F(X)+\alpha G(X)$ splits into linear factors for all rational $\alpha$. However, $F(X)+YG(X)$ is irreducible in $\mathbb Q[Y][X]$, so be Hilbert's irreducibility theorem there are infinitely many $\alpha\in\mathbb Q$ such that $F(X)+\alpha G(X)$ is irreducible. On the other hand, we saw that this polynomial factors into linear factors, so $\deg F\le 1$, contrary to $\deg F\ge2$.

The answer is no. First we see that $T$ is injective: Let $0\ne g\in\mathbb Q[X]$ with $T(g)=0$. Then $\textrm{Gal}(f-\alpha g)'=\textrm{Gal}(f)'$ for all $f\in\mathbb Q[X]$ and $\alpha\in\mathbb Q$. Pick $\beta\in\mathbb Q$ with $g(\beta)\ne0$, and $f\in\mathbb Q[X]$ of degree $n$ larger than $\deg g$ and $2$ such that $\textrm{Gal}(f)=S_n$. Then $\textrm{Gal}(f)'=S_n'=A_n$. But the polynomial $f(X)-\frac{f(\beta)}{g(\beta)}g(X)$ has the root $\beta$, so its Galois group is a subgroup of $S_{n-1}$. But no subgroup of $A_{n-1}=S_{n-1}'$ is isomorphic to $A_n$.

Thus we find $f\in\mathbb Q[X]$ of degree $2$ such that $F=T(f)$ has degree at least $2$. Since $\textrm{Gal}(F)=\textrm{Gal}(f)'=1$, $F$ splits into linear factors. Next pick $\gamma\in\mathbb Q$ such that $T(X+\gamma)=T(X)+\gamma T(1)$ has no common root with $F$. (Recall that $T(1)\ne0$.) Then $F$ and $G=T(X+\gamma)$ are relatively prime. As $f+\alpha(X+\gamma)$ has degree at most $2$ for all rational $\alpha$, we see that $F(X)+\alpha G(X)$ splits into linear factors for all rational $\alpha$. However, $F(X)+YG(X)$ is irreducible in $\mathbb Q[Y][X]$, so by Hilbert's irreducibility theorem there are infinitely many $\alpha\in\mathbb Q$ such that $F(X)+\alpha G(X)$ is irreducible. On the other hand, we saw that this polynomial factors into linear factors, so $\deg F\le 1$, contrary to $\deg F\ge2$.