I have to find the set of $\mathbb{F}_{27}$-Automorphisms of $\mathbb{F}_{19683}$, where $\mathbb{F}_n:=\mathbb{Z}/\mathbb{Z}_n$.
I know that since $27=3^3$ and $19683=3^9$ and $3\vert{9}$, there exists a morphism from $\mathbb{F}_{27}$ to $\mathbb{F}_{19683}$. Also that ${\rm Aut}_{\mathbb{F}_{p}}(\mathbb{F}_{p^n})=\left\langle F_{rp}\right\rangle$ and $\vert{{\rm Aut}_{\mathbb{F}_{p}}(\mathbb{F}_{p^n})}\vert=n$, for a prime number $p$, where $F_{rp}$ is the Frobenius automorphism.
With this I know that ${\rm Aut}_{\mathbb{F}_{3}}(\mathbb{F}_{3^3})=\left\langle F_{r3}\right\rangle$ and ${\rm Aut}_{\mathbb{F}_{3}}(\mathbb{F}_{3^9})=\left\langle F_{r3}\right\rangle$, with cardinalities $3$ and $9$ respectively. So I would have to find the elements of ${\rm Aut}_{\mathbb{F}_{3}}(\mathbb{F}_{3^9})=\left\langle F_{r3}\right\rangle$ that fix $\mathbb{F}_{27}$.
Then I found this post Determine the Automorphisms of the extension field $\mathbb{F}_{19683}$ of $\mathbb{F}_{27}$, so I know I am on the right path. But I don't know how to actually find those elements that fix $\mathbb{F}_{27}$. The only way I can think of is explicitly calculating each of the nine elements of ${\rm Aut}_{\mathbb{F}_{3}}(\mathbb{F}_{3^9})$ and see which fix $\mathbb{F}_{27}$.
I am sorry if this is very trivial, but I am stuck and I don't know what I am missing.
Remember the characterization $$\mathbb{F}_{3^n}=\{ x\in \overline{\mathbb{F}_3}\mid x^{3^n}=x\},$$ namely this is the fixed element set of $\mathrm{Fr}_3^n$, the $n$-th power of the Frobenius map, inside the algebraic closure. It also shows that $\mathrm{Fr}_3$ is of order $n$ on $\mathbb{F}_{3^n}$.
Therefore, the automorphism fixing $\mathbb{F}_{3^3}$ is just the power $\mathrm{Fr}_3^3\colon x\mapsto x^{27}$, which generates the subgroup of order 3 inside the group $\mathrm{Aut}_{\mathbb{F}_3}(\mathbb{F}_{3^9})=\langle \mathrm{Fr}_3\mid \mathrm{Fr}_3^9=1\rangle$.