Question
Let's suppose I have two polynomials $f \in F[x]$ and $g \in F[x]$ such that:
i) $\mathbb{Q} \subseteq F$;
ii) the roots of $f$ belong to a field $K$;
iii) and their roots satisfy:
$r_g = \displaystyle\sum_i \left[ q_i \cdot r_{fi} \right]$
where $r_g$ and $r_f$ represents, respectively, the roots of $g$ and $f$; and $q_i$ is a rational number. Since their roots satisfy the equation above, $f$ and $g$ share the same splitting field $K$. Are the Galois Groups of $f$ and $g$ the same?
I.e., is it enough to say they share the same Galois Group?
I think so, but I cannot come up with a rigorous (even a good) argumentation.
As the comments have explained: if $K$ is both the splitting field of $f$ and of $g$ then the Galois group of both polynomials is $G$, the set of automorphisms of $K$ which fix $F$ elementwise.
However (and it is a big However) if we are interested not just in the abstract properties of $G$ but in how it acts on the sets of roots of $f$ and $g$ then these actions can be very different.
As an example take $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ which is the splitting field of both $f(X):=(X^2-2)(X^2-3)$ and $g(X):=X^4-22X^2+49$. Acting on the roots of $f$ we have that $G=\{e, (12), (34), (12)(34)\}$; acting on the roots of $g$ we have that $G=\{e, (12)(34), (13)(24), (14)(23)\}$. The subgroups of $S_4$ are isomorphic, but not conjugate.
In fact polynomials $f$ and $g$ with the same splitting field and Galois group need not even be of the same degree. Amongst others there are quartics and sextics which have the same splitting field.