Statement: $K(X)/K(X^n)$ is Galois if and only if $p$ doesn't divide $n$ and $K$ contains all the zeroes of the separable polynom $g := X^n-1 \in K[X]$. (the polynom is certainly separable because $p$ doesn't divide $n$, so the roots of $g$ is unique.) For simplicity we denote from now on $E$ the set of all the roots of $g$.
But I have doubts about the statement, I can see how the direction (<-) goes: $K(X)$ = $K(X^n)(F)$, $F$ is the zeroes of a separable polynom $f := Z^n-X^n \in K(X^n)[Z]$ ($f$ is separable since $eX$ is a root of $f$ for all $e \in E$ and $|E| = n$) so the extension $K(X^n)(F)/K(X^n)$ is Galois.
For the other direction (->) we assume $p$ divides $n$. Let $n =mp^r$ where $r \geq 0$ and $ggT(p,m) =1$. Then $f$ is no longer separable since we have $f = Z^{mp} -X^{mp} = (Z^m-X^m)^p.$ And I don't know how to proceed from here, is it possible that the statement is not true?
Maybe it can help if we can find the minimal polynom of $X$ in $K(X^n)$? is it $Z^n-X^n$? (assuming $p$ doesn't divide $n$)
Update:
To prove (->): Assume $p$ divides $n$ Let $n =mp^r$ where $r \geq 0$ and $ggT(p,m) =1$. So we have $f = Z^n-X^n = (Z-e_1X)^p ... (Z-e_mX)^p$. Since the extension is Galois, we know that the minimal polynom $\mu_{X,K(X)/K(X^n)} \in K(X^n)[Z]$ is separable and it divides $f$ so it can take at most $m$ factors of $f$, since otherwise we would have doubles. This shows that the minimal polynom would have terms $X^i$ in it with $i \leq m < n$, hence it cannot be in $K(X^n)$, a contradiction. Therefore $p$ cannot divide $n$.
We now know that $p$ doesn't divide $n$, hence $|E|=n$. Furthermore, we know $h:= \mu_{X,K(X)/K(X^n)}$ divides $f$, which in this case can be written as $(Z-e_1X)...(Z-e_nX) \in \overline{K}(X)[Z]$ with $e_i \in E \subset \overline{K}$, we say $h$ must be equal to $f$, then otherwise we would again have terms $X^i$ with $i<n$ in it, which should never happen. Since the extension is Galois, we have $h = \prod_{b \in Gal(K(X)/K(X^n)).X}(Z-b)$, hence $e_iX \in K(X)$ and $e_i \in K(X)$, but $\overline{K}$ was constructed independently of $X$, therefore $e_i$ cannot be expressed in terms of $X$, therefore $e_i$ is in $K$ for all $i$, i.e $K$ contains all the zeroes of $g$, as required.
Apologies for earlier confusions. Let $p$ be the characteristic of $K$ and let the minimal polynomial for $X$ in $K(X^n)[Z]$ be $f'$. Then $f'$ divides $Z^n-X^n$, but since $X$ is a transcendental element over $K$ it's clear that the degree of the extension is $n$, hence $f'=f$ by counting degrees. So $f$ is irreducible and is the minimal polynomial of $X$. The first bit of this is essentially a rephrasing of what you've already said.
Assume $p$ doesn't divide $n$ and that $K$ contains all roots of $X^n-1$. This polynomial is separable - $nX^{n-1}\neq0$, given $p$ doesn't divide $n$. Since $K$ contains all the roots, it is clear that all the roots must then be distinct (a repeated root would have to satisfy $a^{n-1}=0$, so $a=0$, but then $a^n-1=-1\neq0$). So we have $\zeta_1,\zeta_2,\cdots,\zeta_n$, $n$ distinct roots of $X^n-1$ in $K$.
The polynomial $f:=Z^n-X^n$ is separable (since $p$ doesn't divide $n$) and it factors fully in $K(X)[Z]$ as $(Z-\zeta_1X)\cdots(Z-\zeta_nX)$. As $X$'s minimal polynomial in $K(X^n)[Z]$ is $f$, we then know the extension is normal and separable, hence Galois.
Now for the converse.
Assume the extension is Galois. The minimal polynomial $f$ must then be separable and split fully into linear factors in $K(X)[Z]$. $f$ is separable iff. the formal derivative $nZ^{n-1}$ is nonzero, and this is true iff. $p$ doesn't divide $n$. As $f$ splits fully into linear factors, write: $$Z^n-X^n=(Z-\alpha_1)\cdots(Z-\alpha_n)$$For distinct $\alpha_i\in K(X)$, $\alpha_1=X$. Define $\zeta_i:=\alpha_i\cdot X^{-1}$. Then $\zeta_i^n=1$ for every $i$. We know each $\zeta_i$ is distinct and is an element of $K(X)$, but we further want to know they are elements of $K$.
It is a general fact that any element of $K(X)$ algebraic over $K$ must be in $K$, however.
Let $\gamma\in K(X)$ satisfy $h(\gamma)=0$ where $h\in K[Z]$. Let $F$ be an algebraic closure of $K$. I know $h$ splits into linear factors in $F$ and all roots are elements $c_1,c_2,\cdots,c_k$ of $F$. Consider the roots of $h$ in $K(X)$: the embedding $K\hookrightarrow F$ ensures these are also all roots in $F(X)$. But we know all the roots of $h$ in $F$ are constant elements, so all the roots of $h$ in $F(X)$ are constant elements (no "$X$" terms). So it follows that the roots of $h$ in $K(X)$ were also all constant elements, in $K$, as desired. Hence $\gamma\in K$.