You correctly state that the normal closure of $F = \mathbb{Q}(\sqrt[5]{3})$ is $L = \mathbb{Q}(\sqrt[5]{3}, \zeta)$, where $\zeta$ is a primitive $5^\text{th}$ root of unity. $L$ has two important subfields, $F$ and $K = \mathbb{Q}(\zeta)$. Note that $K/\mathbb{Q}$ is Galois, since the other roots of its minimal polynomial $\frac{x^5 - 1}{x-1} = x^4 + x^3 + x^2 + x +1$ are just the powers of $\zeta$. Also note that $[\mathbb{Q}(\zeta): \mathbb{Q}] = \varphi(5) = 4$ and $[\mathbb{Q}(\sqrt[5]{3}) : \mathbb{Q}] = 5$ are relatively prime, so $20$ divides $[\mathbb{Q}(\sqrt[5]{3}, \zeta) : \mathbb{Q}]$. Since $[\mathbb{Q}(\sqrt[5]{3}, \zeta) : \mathbb{Q}] \leq 20$, this shows that $[\mathbb{Q}(\sqrt[5]{3}, \zeta) : \mathbb{Q}] = 20$. Thus we have the following field diagram
$\hspace 2.5cm$
and the corresponding Galois group diagram.
$\hspace 2cm$
Since $K/\mathbb{Q}$ is Galois, then $N = \text{Gal}(L/K)$ is normal in $G = \text{Gal}(L/\mathbb{Q})$. Moreover, since $N \cap H = 1$ by order considerations, then $|NH| = \frac{|N||H|}{|N \cap H|} = \frac{4 \cdot 5}{1} = 20$, so $NH = G$. Then $G \cong N \rtimes H$, i.e., $G$ is the semidirect product of $N$ and $H$. Since $K \cap F = \mathbb{Q}$ and $L = KF$, then
$$
H = \text{Gal}(L/F) \cong \text{Gal}(K/\mathbb{Q}) = \text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}) \cong (\mathbb{Z}/5\mathbb{Z})^\times \cong \mathbb{Z}/4\mathbb{Z}
$$
by results on composite fields and cyclotomic extensions. Now $|N| = 5$ so $N \cong \mathbb{Z}/5\mathbb{Z}$, hence we have (abstractly), $G \cong N \rtimes H \cong \mathbb{Z}/5\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$.
Maybe you're satisfied with that, but we can also get a more concrete description by examining the action of $H$ on $N$ in the semidirect product. Let
\begin{align*}
\sigma: L &\to L\\
\sqrt[5]{3} &\mapsto \sqrt[5]{3}\\
\zeta &\mapsto \zeta^2
\end{align*}
\begin{align*}
\tau: L &\to L\\
\sqrt[5]{3} &\mapsto \zeta\sqrt[5]{3}\\
\zeta &\mapsto \zeta
\end{align*}
Note that $\sigma$ fixes $F$ and $\tau$ fixes $K$, so $\sigma \in H$ and $\tau \in N$. Moreover, since $2$ generates $(\mathbb{Z}/5\mathbb{Z})^\times$, then $\sigma$ generates $H$, and since $\tau$ has order $5$, it generates $N$. Since $N \trianglelefteq G$, then $H$ acts on $N$ by conjugation. Note that $\sigma^{-1}$ is $\zeta \mapsto \zeta^3$ since $3 = 2^{-1} \pmod{5}$. Computing $\sigma \tau \sigma^{-1}$, we find
\begin{align*}
\sigma \tau \sigma^{-1}: \sqrt[5]{3} &\overset{\sigma^{-1}}{\longmapsto} \sqrt[5]{3} \overset{\tau}{\longmapsto} \zeta \sqrt[5]{3} \overset{\sigma}{\longmapsto} \zeta^2 \sqrt[5]{3}\\
\zeta &\overset{\sigma^{-1}}{\longmapsto} \zeta^3 \overset{\tau}{\longmapsto} \zeta^3 \overset{\sigma}{\longmapsto} \zeta \, .
\end{align*}
This is exactly the same action as $\tau^2$, so $\sigma \tau \sigma^{-1} = \tau^2$, i.e., $\sigma \tau = \tau^2 \sigma$, which provides us with a commutation relation. This allows us to write every element of $G$ as $\tau^i \sigma^j$ for some $i \in \{0, 1, 2, 3\}$ and $j \in \{0, 1, 2, 3, 4\}$, which accounts for all $20$ of the elements of $G$. Thus we have the presentation
$$
G = \langle \sigma, \tau \mid \sigma^4 = \tau^5 = 1, \sigma \tau = \tau^2 \sigma \rangle \, .
$$
As you can see here, this is a presentation for the Frobenius group of order $20$.
