= Solution
An automorphism sends the <primitive root of unity> $\xi_n$ to another <primitive root of unity>, hence uniquely to $\xi_n^a$ with $a\in(\mathbb Z/n\mathbb Z)^*$. Define $\chi(\sigma)=a$. Composition multiplies exponents, and an automorphism with exponent one fixes the generator and thus all of $L$. Therefore \b[$\chi$ is an injective group homomorphism] and $G$ is abelian.
The extension is finite Galois: it is the <splitting field> of $X^n-1$, whose roots are all present and distinct. Existence of a root of order $n$ in positive characteristic implies that the characteristic does not divide $n$. The <Fundamental theorem of Galois theory> identifies an intermediate field with $L^H$. Since $G$ is abelian, $H$ is normal; the normal-subgroup criterion gives \b[$L^H/K$ Galois], with group $G/H$.
For a non-surjective nontrivial example, take $K=\mathbb Q(i)$ and $n=8$. Its cyclotomic extension has degree two, and its image is $\{1,5\}$ inside the four-element group $(\mathbb Z/8\mathbb Z)^*$. For $K=\mathbb Q$ and prime $p$, $\Phi_p(X)=1+\cdots+X^{p-1}$ is irreducible: $\Phi_p(X+1)$ is Eisenstein at $p$. Thus the extension has degree $p-1$, equal to the <unit group>'s order, proving surjectivity.
For $n=7$, let $\zeta=\xi_7$. The cyclic <Galois group> of order six has exactly one subgroup of each order dividing six, so there are exactly four fields:
$$
\begin{array}{c|c}
M&\operatorname{Aut}(\mathbb Q(\zeta)/M)\\\hline
\mathbb Q&\{\sigma_a:a=1,2,3,4,5,6\}\\
\mathbb Q(\sqrt{-7})&\{\sigma_1,\sigma_2,\sigma_4\}\\
\mathbb Q(\zeta+\zeta^{-1})&\{\sigma_1,\sigma_6\}\\
\mathbb Q(\zeta)&\{\sigma_1\}
\end{array}
$$
Here $\sigma_a(\zeta)=\zeta^a$. To identify the quadratic field without guessing, put $A=\zeta+\zeta^2+\zeta^4$ and $B=\zeta^3+\zeta^5+\zeta^6$. Multiplication gives $A+B=-1$ and $AB=2$, so $(A-B)^2=-7$. The order-three subgroup fixes $A$, and $A-B\notin\mathbb Q$, proving the quadratic identification. The real field is cubic because complex conjugation is $\sigma_6$, with fixed subgroup of order two.
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