Prove that $\mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$ is a subgroup of $\mathrm{Aut}(L/K)$

Let $L/K$ be a Galois extension of fields and let $\zeta_n$ be a primitive $n$th root of unity in some field extension of $L$, where $n$ is not divisible by the characteristic. Prove that $\mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$ is a subgroup of $\mathrm{Aut}(L/K)$.

I'm unsure because I don't really know what is required. If $\sigma \in \mathrm{Aut}(L(\zeta_n)/K(\zeta_n))$, then $\sigma$ maps $\zeta_n$ to another primitive $n$th root of unity. Why must it restrict to an automorphism of $L$?

Thanks

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It's interesting that you have a question about your own theory. :) Search on the term "theorem of natural irrationalities" for the general issue you ask about. If $L$ and $F$ are finite extensions of $K$ and $L/K$ is Galois, then $LF/F$ is a finite Galois extension and ${\rm Gal}(LF/F)$ can be naturally identified with a subgroup of ${\rm Gal}(L/K)$ by restricting automorphisms to $L$. – KCd May 15 '12 at 1:44

Because $L/K$ is a Galois extension, hence it is normal, so $\sigma$ must map $L$ into itself.