= Solution
The <cyclotomic polynomial> $\Phi_7=1+X+\cdots+X^6$ is irreducible: replacing $X$ by $X+1$ gives an <Eisenstein polynomial> at seven. Hence the extension has degree six, and
$$
\boxed{G=(\mathbb Z/7\mathbb Z)^\times\cong C_6,\qquad\sigma_a(\eta)=\eta^a.}
$$
There is one subgroup of each order $1,2,3,6$, and therefore precisely four intermediate fields. The trivial subgroup fixes $\mathbb Q(\eta)$; the whole group fixes $\mathbb Q$. The order-two subgroup $\{1,6\}$ is <complex conjugation> and fixes $\mathbb Q(\eta+\eta^{-1})$, of degree three: $\eta$ satisfies $X^2-(\eta+\eta^{-1})X+1$ and is not real.
The order-three subgroup $\{1,2,4\}$ fixes $u=\eta+\eta^2+\eta^4$. Its other conjugate is $v=\eta^3+\eta^5+\eta^6$. Expanding products and using $\sum_{j=1}^6\eta^j=-1$ gives $u+v=-1$ and $uv=2$. Thus $u^2+u+2=0$, and this <fixed field> is $\mathbb Q(u)=\mathbb Q(\sqrt{-7})$, of degree two. These generators have the required degrees and are fixed by the indicated subgroups, so the <Galois correspondence> proves they are the full fixed fields.
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