Let S be a finite field of characteristics 2 and the map be define as
Show that $\eta$ is automorphism, i.e., S is isomorphism of itself.
Proof: Now for one there is the trivial automorphism x$\longmapsto$x, and by composing this so-called Frobenius automorphism with itself multiple times which is x$^p$. Let us take p=2. Therefore it is not an automorphism.
I would like have I proved this correctly?
I am lost to prove that the ring $\mathbb Q$(i)[x] with order of 2 and 4 is automorphism.