Let $E = \mathbb{Q}(\omega_7)$, where $\omega_7$ is the 7th root of unity. We know that $$\mathbb{Q}(\omega_7) \cong \mathbb{Z}_7^{\times},$$ where $\mathbb{Z}_7$ is the multiplicative group of units, given by the elements coprime to $7$. Therefore $\mathbb{Z}_7 = \{1,2,3,4,5,6 \}$. It therefore follows that $\left| \text{Gal}(E/\mathbb{Q}) \right| = 6$ and the minimal polynomial of $\omega_7$ is 6. Since 7 it is prime, we know that the minimal polynomial of $\omega_7$ is given by $$x^n-1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1),$$ i.e. the minimal polynomial is $x^6+x^5+x^4+x^3+x^2+x+1$. Which has degree 6, verifying what was found above. Furthermore, a basis for $E/\mathbb{Q}$ is given by $\{1, \omega, \omega^2, \omega^3, \omega^4, \omega^5,\omega^6 \}$. We know that the automorphisms of the Galois group are given by \begin{align*} \omega_7 \mapsto \omega_7 && \omega_7 \mapsto \omega_7^2 && \omega_7 \mapsto \omega_7^3 \\ \omega_7 \mapsto \omega_7^4 && \omega_7 \mapsto \omega_7^5 && \omega_7 \mapsto \omega_7^6. \end{align*} Note that the entire field extension is fixed by the first automorphism, so we wont concern ourselves with that.
How do we proceed from here to find the intermediate field extensions and their corresponding subgroup of $\mathbb{Z}_7^{\times}?$
As noted, $\operatorname{Gal}(\mathbb{Q}(\omega_7)/\mathbb{Q}) \cong \mathbb{Z}^\times_7 \cong \mathbb{Z}_6$. One can check that
$$ \omega_7 \mapsto \omega_7^3 $$
generates the Galois group. Since
$$ 3^2 \equiv 2 \mod 7, \quad 3^3 \equiv 6 \mod 7, $$
$$ \omega_7 \mapsto \omega_7^2, \quad \omega_7 \mapsto \omega_7^6 $$
generate subgroups of the Galois group of orders 3 and 2 respectively. Note that the second automorphism is just complex conjugation. By the Galois correspondence, the fixed fields of these automorphisms are precisely the intermediate fields between $\mathbb{Q}(\omega_7)$ and $\mathbb{Q}$, of degrees 3 and 2 over $\mathbb{Q}$, respectively.