Problem 1.
Let S be a finite field of characteristics 2 and the map be define as
$\eta$: S$\longrightarrow$S
x$\longmapsto$x$^p$
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?
Problem 2
I am lost to prove that the ring $\mathbb Q$(i)[x] with order of 2 and 4 is automorphism.
Your question isn't exactly worded the best, so I'm confused about what precisely it is that you are asking. However, the Frobenius automorphism is indeed an automorphism:
Let $p=2$. Certainly, $\eta(0) = 0^2 = 0$.
$\eta(xy) = (xy)^2 = x^2y^2$ since fields are commutative.
$\eta(x+y) = (x+y)^2 = x^2+2xy+y^2 = x^2+y^2 = \eta(x)+\eta(y)$ since $2xy \equiv$ $0$ (mod 2).
We conclude $\eta:S \rightarrow S$ is indeed an automorphism.
Unfortunately, I'm rather unsure what Problem 2 is asking.