Let $N=:\mathbb{Q}(\mathcal{R}_{x^5-100})$, $L=:\mathbb{Q}(\zeta_5)$, where $\zeta_5$ is a primitive $5^{th}$-root of unity, and $\Gamma:=\text{Aut}(N|\mathbb{Q})$.
1) Prove that $\text{Aut}(N|L)<\Gamma$ is cyclic with order $5$ and is a normal subgroup.
2) Prove that $\text{Aut}(N|\mathbb{Q}(\sqrt[5]{100}))<\Gamma$ is cyclic with order $4$ and is not a normal subgroup.
I already know that $\text{Aut}(N|L)$ must be cyclic, whose order divides $5$. Since $N|L$ is clearly non trivial, then $|\text{Aut}(N|L)|=5$.
I can also prove that $[N:\mathbb{Q}]=20$, so that $[N:\mathbb{Q}(\sqrt[5]{100})]=4$, and that $\text{Aut}(N|\mathbb{Q}(\sqrt[5]{100}))$ must be cyclic.
But I'm having trouble to prove normality in 1) and non-normality in 2), because I can't exhibit the automorphisms. For example, what are the possibilities for $\sigma\in\text{Aut}(N|L)$? Since $N$ is generated by $\{1, \sqrt[5]{100}, ..., (\sqrt[5]{100})^{4}\}$ as an $L$-vector space, I know that $\sigma(\sqrt[5]{100})=(\sqrt[5]{100})^k$ where $k=1, 2, 3$ or $4$ are possibilities for $\sigma$. But there is still one possibility missing, and I can't find it.
I can't exhibit the automorphisms of $\text{Aut}(N|\mathbb{Q})$ either, and it seems complicated. How am I supposed to tackle this problem? Thanks!
Firstly you haven't proven cyclicity for the second case since you have only confirmed the order. To see it is cyclic, note that the automorphisms:
$$\begin{cases} \sqrt[5]{100}\mapsto\sqrt[5]{100} \\ \zeta_5\mapsto \zeta_5^k && 1\le k\le 4 \end{cases}$$
work for those purposes. To see one is normal and the other isn't, just use the fundamental theorem of Galois theory: since $\Bbb Q(\zeta_5)$ is Galois and $\Bbb Q(\sqrt[5]{100})$ is not, the corresponding stabilizer subgroups have the corresponding normality properties.
Also, your candidate automorphisms are a bit off: $\text{Aut}(N|L)$ is composed of
$$\sigma_k:\begin{cases}\sqrt[5]{100}\mapsto\zeta_5^k\sqrt[5]{100} && 0\le k\le 4 \\ \zeta_5\mapsto \zeta_5 \end{cases}$$