Compute Galois group of $\mathbb F_q/\mathbb F_p$

There is a bit of confusion in your attempt but some ideas are alright. Let me try to set this straight.

The degree of the extension is $n$. Thus the Galois group has $n$ elements.

You know one, the Frobenius automorphism $\sigma$. Now, for $k < n$ (not $p^n$) one has that $\sigma^{k}(x) = x^{p^k}$ (not $x^{pk}$, note that $\sigma(\sigma(x))= \sigma(x^p)=(x^p)^p = x^{p^2}$ and so on), and then, since $x^{p^k} \neq x$ (not $1$) one concludes $\sigma^k$ is not the identity map (that is it is not the "one-element" of the Galois group).

Consequently the order of $\sigma$ is (at least) $n$, and the $n$ automorphisms $\sigma^{0}, \sigma^1, \dots, \sigma^{n-1}$ are all distinct. Thus this must be the full Galois group. That is, the Galois group is cyclic of order $n$; more precisely, it is generated by the Frobenius automorphism.