Simple field extension and roots of a polynomial

Question

Let $K$ be a field, $f \in K[X]$ separable and irreducible with $\text{deg}(f)=n$; $x_1,...,x_n$ are the roots of $f$ in a splitting field of $f$ over $K$. Let $g \in K[X]$ be any polynomial with $\text{deg}(g)\leq n-1$. I want to prove the following statement: $$ K[g(x_1)]=K[x_1] \Longleftrightarrow g(x_1),...,g(x_n) \ \text{are pairwise distinct}.$$

Answer 1

Hint 1:

$K(g(x_1)) \subset K(x_1)$ is always true, so the two are equal iff $[K(x_1):K]=[K(g(x_1)):K]$.

Hint 2:

In any finite Galois extension $L/K$, if $G = Gal(L/K)$, the minimal polynomial of $y$ over $K$ is $f(X)=\Pi_{i=1}^r(X-y_i)$ where $\{y_1,\dots,y_r\}$ is the orbit of $G$ acting on $y$, and the $y_i$ are distinct.* Since $f(X)$ is the minimal polynomial of $x_1$, if $L$ is the splitting field for $f(X)/K$, $G=Gal(L/K)$, then $G$ acts transitively on the roots of $f$ by what I said above. What does this tell us about the minimal polynomial of $g(x_1)$?

*This is because $f(X)$ is fixed by $G$ so it's coefficients lie in $K$, and any polynomial in $K[X]$ which has $y$ as a root has all of the $y_i$ as roots as well.