Galois Group for $x^5-1$

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.

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.