Prove that the Galois group of $x^n-1$ is abelian over the rationals

Your argument up to the point where you're stuck seems completely correct to me.

To finish it, suppose that $\sigma(\omega) = \omega^j$ and that $\tau(\omega) = \omega^k$ for some $j,k \in \mathbb{N}$. Then just observe that $$ (\sigma \circ \tau)(\omega) = \sigma(\tau(\omega)) = \sigma(\omega^k) = \sigma(\omega)^k = (\omega^j)^k = \omega^{jk} $$ and similarly $$ (\tau \circ \sigma)(\omega) = \tau(\sigma(\omega)) = \sigma(\omega^j) = \tau(\omega)^j = (\omega^k)^j = \omega^{jk} $$

You are correct that the group of all permutations of roots of unity is not abelian. But not all permutations actually give a field automorphism.

The important fact is this: once you know $\sigma(\omega)$, you also know $\sigma(\omega^2)$, $\sigma(\omega^3)$, etc.