How to find an automorphism to enumerate the subfields of this cyclotomic field? Here, $\zeta$ denotes the primitive $9$-th root of unity.  Since the minimal polynomial of $\zeta$ over $\mathbb Q$ is $x^6 + x^3 + 1$, the field extension $\mathbb Q(\zeta) / \mathbb Q$ has degree $6$.

*

*The full set of roots is $\{ \, \zeta^i : i = 1, 2, 4, 5, 7, 8 \, \}$.

*On the other hand, a basis of $\mathbb Q(\zeta)/\mathbb Q$ is $\{ \, 1, \zeta, \zeta^2, \zeta^3, \zeta^4, \zeta^5 \, \}$.

This is where I am getting confused.  Since $\mathbb Q$-automorphisms permute the roots of the minimal polynomial, to find a fixed field associated with a subgroup of automorphisms, I should pick something like $\sigma \in \text{Aut} (\mathbb Q(\zeta)/\mathbb Q)$ given by:
$$ \zeta \mapsto \zeta^2 \mapsto \zeta^4 \mapsto \zeta^5 \mapsto \zeta^7 \mapsto \zeta^8 \mapsto \zeta$$
So I am permuting the elements in bullet 1.  Clearly, $\{ \, 1, \sigma^2, \sigma^4 \, \}$ is a subgroup of $\text{Aut } \mathbb Q(\zeta)$ with order $3$.  To see what an element fixed by $\sigma^2$ looks like, we have:
$$a + b\zeta + c\zeta^2 + d\zeta^3 + e\zeta^4 + f\zeta^5 \mapsto ???$$
but $1, \zeta^3$ are not in the list of elements associated $\sigma$.  What did I misunderstand?
 A: Your misunderstanding is in the structure of the automorphisms of $\Bbb{Q}(\zeta)$. There is no field automorphism $\sigma\in\operatorname{Aut}\Bbb{Q}(\zeta)$ permuting the powers of $\zeta$ in the way you describe:
If $\sigma(\zeta)=\zeta^2$ then $\sigma(\zeta^4)=\sigma(\zeta)^4=(\zeta^2)^4=\zeta^8$. In fact, in this way all values of $\sigma(\zeta^k)$ can be computed given that $\sigma(\zeta)=\zeta^2$ alone. The orbits
$$\sigma:\ \zeta\ \mapsto\ \zeta^2\ \mapsto\ \zeta^4\ \mapsto\ \zeta^8\ \mapsto\ \zeta^7\ \mapsto\ \zeta^5\ \mapsto\ \zeta,$$
$$\sigma:\ \zeta^3\ \mapsto\ \zeta^6\ \mapsto\ \zeta^3\qquad\text{ and }\qquad\sigma:\ 1\ \mapsto\ 1,$$
show that $\sigma$ acts transitively on the roots of $x^6+x^3+1$, and because these powers of $\zeta$ span $\Bbb{Q}(\zeta)$ over $\Bbb{Q}$, it shows that $\sigma(\zeta)=\zeta^2$ determines $\sigma$ entirely. You should check that $\sigma$ does indeed extend to a field automorphism of $\Bbb{Q}(\zeta)$.
This argument shows that every $\tau\in\operatorname{Aut}\Bbb{Q}(\zeta)$ is determined by $\tau(\zeta)$ alone, which must be in
$$\{\zeta,\zeta^2,\zeta^4,\zeta^5,\zeta^7,\zeta^8\}.$$
Check that all these values for $\tau(\zeta)$ extend to field automorphisms of $\Bbb{Q}(\zeta)$, thus yielding a bijection
$$\operatorname{Aut}\Bbb{Q}(\zeta)\ \longleftrightarrow\ (\Bbb{Z}/9\Bbb{Z})^{\times}.$$
Next check that this is a group homomorphism, and hence an isomorphism. Now enumerating the subgroups of $\operatorname{Aut}\Bbb{Q}(\zeta)$ is very easy. For finding the corresponding fixed subfields you seem to have the right idea already, so give that another try once the above makes sense.
Do note that this also shows that arbitrary permutations of the powers of $\zeta$ in general do not extend to field automorphisms.
