Determine the Automorphisms of the extension field $\mathbb{F}_{19683}$ of $\mathbb{F}_{27}$ I am trying to find $\text{Aut}_{\mathbb{F}_{27}}(\mathbb{F}_{19683})$. I have a theorem that looks promising but can not make it work.
For a field $\mathbb{K}$ with $\text{char}(\mathbb{K})=p$ prime we have the Frobeniusmorphism $\text{Fr}_p:\mathbb{K} \rightarrow \mathbb{K}, \ \alpha \mapsto \alpha^p$. Then $(\text{Fr}_p)^n$ is also a morphism. The theorem I am trying to use states that
$\text{Aut}_{\mathbb{F}_{p}}(\mathbb{F}_{p^n}) = \langle Fr_p \rangle$.
Obviously my problem is that 27 is not prime. Rewriting the problem $\text{Aut}_{\mathbb{F}_{3^3}}(\mathbb{F}_{3^{3^3}})$ and the fact that the theorem was mention just above this exercise make me think that one can make use of it to solve the problem.
Any help appreciated.

Note that for general field extensions $E/F/K$ we have a restriction homomorphism $$\psi:\operatorname{Aut}_K(E)\to\operatorname{Aut}_K(F)$$
with kernel $\operatorname{Aut}_F(E)$. Applying this to the situation $E=\mathbb{F}_{19683},F=\mathbb{F}_{27},K=\mathbb{F}_{3}$ we see that the group $\operatorname{Aut}_F(E)$ is exactly that subgroup of $\langle Fr_3\rangle=\operatorname{Aut}_K(E)$ which fixes $\mathbb{F}_{27}$. Since $\langle Fr_3\rangle$ is cyclic it shouldn't be too hard to find a generator for that.
