Galois Group for $x^5-1$ This question is an extension to the question in math.stackexchange.com/questions/759230/subfield-of-the-galois-group-of-x5-1
It seems the discussion in that topic is dead and I still have a major question. 
For the $\Phi_5(x)$ the complex roots are $\{\zeta, \zeta^2, \zeta^3, \zeta^4 \}$ for $\zeta:=e^{\frac{2\pi i}5}$. In his response Don Antonio defined $\;\omega\in Gal(\Bbb Q(\zeta)/\Bbb Q)\;$ to be complex conjugation.
Don went on to explain: 
for$\;z\in\Bbb C\;\;,\;\;\overline z=z^{-1}\iff |z|=1\;$ , and thus:
$$\omega(\zeta+\zeta^{-1})=\omega(\zeta)+\omega(\zeta)^{-1}=\zeta^{-1}+\zeta\implies \zeta+\zeta^{-1}\in\Bbb R$$
end of quote
Now Let $Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) = \{id, \rho, \rho^2, \rho^3\}$  where $\rho(\zeta) = \zeta^2$ and $\rho^2(\zeta) = \zeta^3$ and $\rho^3(\zeta) = \zeta^4$ and so on.
Could someone write this complex conjugation $\omega$ in terms of the automorphism $\rho$
I tried all kinds of conjugations and cannot find the one that will send
 $\zeta \rightarrow \zeta^4, \zeta^2 \rightarrow \zeta^3$
 A: The Galois group for any cyclotomic field $\mathbb{Q}(\zeta_n)$ will be the multiplicative group $\mathbb{Z}_n^{\times}$.  When $n$ is prime, as is the case here, then $\mathbb{Z}_n^{\times} \cong \mathbb{Z}_{n-1}$, which is cyclic.  (The latter fact can be proven using the structure theorem for finitely generated abelian groups.)
Therefore, we simply need to find a generator for the Galois group.  To do this, note that every $\mathbb{Q}$-automorphism of this field is determined by its action on $\zeta$, and further, $\zeta \mapsto \zeta^k$ always defines a legitimate automorphism (since $\text{Gal}(f)$ acts transitively on the roots of $f$ when $f$ is irreducible).  Therefore, I simply experiment as follows:
Does $\phi$ such that $\phi(\zeta) = \zeta^2$ generate the group?  If not, what about $\phi(\zeta) = \zeta^3$?  And I continue until I have found my generator -
 it'll end up being $\zeta \mapsto \zeta^k$ for a $k$ that generates $\mathbb{Z}_n^\times$.  
Once you've found the generator $\phi$, the Galois group is simply $\{ \text{id}, \phi$, $\phi^2$, $\phi^3\}$.  Noting that $\mathbb{Z}_4$ has one subgroup isomorphic to $\mathbb{Z}_2$, you will get a subgroup $\{ \text{id}, \phi^2\}$ that sends $\zeta \mapsto \zeta^4$ and $\zeta^2 \mapsto \zeta^3$ when it acts on the roots.  
Long story short, I think this is simpler than trying to think about the automorphisms in terms of complex conjugation, etc.
A: If $\rho^3(\zeta)=\zeta^4$ then $$\rho^3(\zeta^2)=(\rho^3(\zeta))^2=(\zeta^4)^2=\zeta^8=\zeta^3,$$
which shows that $\omega=\rho^3$. Your notation is rather misleading though; if $\rho(\zeta)=\zeta^2$ then 
$$\rho^2(\zeta)=\rho(\zeta^2)=(\rho(\zeta))^2=(\zeta^2)^2=\zeta^4.$$
You may want to consider renaming the elements of the Galois group.
