Invariant subspaces of Frobenius automorphism Given finite field, is it true that any invariant subspace of its
  Frobenius automorphism is a subfield?


Firstly, I know that $F_{p^k}$ is a subfield of $F_{p^n}$ if and only if $k$ divides $n$.

For the second, it seems that we shall use that if finite field is of characteristic $p$ then any element satisfies $x^p=x$.

However, I can't proceed to the final answer from this remarks. Can anyone explain me is it truth or not and why?
$\newcommand{\F}{\mathbb{F}}$I understand the question to be


  Suppose the $\F_{p}$-subspace $V$ of the finite field $\F_{p^{n}}$ is invariant under the Frobenius automorphism. 
  Is $V$ a subfield of $\F_{p^{n}}$?


The answer is negative.

Let $E = \F_{p^{6}}$. 

Then $U = \F_{p^{2}}$, $W = \F_{p^{3}}$ are subfields of $E$, and thus $\F_{p}$-subspaces of $E$. 

Then $V = U + W$ is an $\F_{p}$-subspace of $E$, which has dimension $2 + 3 - 1 = 4$ over $\F_{p}$, and thus order $p^{4}$, so that it is not a subfield of $E$. 

However $V$ is invariant under the Frobenius automorphism, as $U$ and $W$ are.
It seems that I found a proof. Denote our finite field of characteristic $p$ by $F_{p^n}$. Then set of all elements invariant under $x \mapsto x^p$ contains prime subfield $F_p$ and roots of the polynomial $t^p-t$, but $\vert F_p \vert = p$, so the invariant subset is exactly the prime subfield $F_p$ and the answer is yes.

Moreover, multiplication group $F_{p^n}^*$ is cyclic of order $p^n-1$ and by Lagrange theorem any its element has order $k \in \mathbb{N}$, where $k$ divides $p^n-1$.

Is everything correct?