Why $f$ is invariant for $G_f$. Let $F$ a field and $f\in F[X]$ a separable polynomial. Let $K_f$ the splitting field of $f$ and $G_f=Gal(K_f/F)$ its Galois group. We suppose it act transitively on the roots of $f$.$\alpha _1,...,\alpha _n$ the roots of $f$. Then $$f=\prod_{i=1}^n(X-\alpha _i)\in K_f[X].$$
In my solution, it's written that $f$ is invariant by $G_f$ (i.e. $\sigma (f)=f$ for all $\sigma \in G_f$), But to me $$\sigma (f(X))=\sigma \left(\prod_{i=1}^n (X-\alpha_i) \right)=\prod_{i=1}^n(\sigma (X)-\alpha _i)=f(\sigma (X))$$
since $G_f$ act transitively on the roots of $f$. So finally, we don't have $\sigma (f(X))=f(X)$, so what's wrong in my argument ? May be my teacher is wrong and $f$ is not a fix point of $\sigma $, what do you think ?

Note that the standard way for the automorphisms $\sigma$ of the field $K_f$ being extended to the polynomial ring $K_f[X]$ is precisely by imposing that $\sigma(X)= X$. It is then usually also denoted by $\sigma$. Thus $\sigma(X)= X$ by definition. 
