Let $k$ be a perfect field of characteristic $p > 0$ and let $X \to S = \textrm{Spec } k$ be a scheme. Let $F : X \to X$ be its absolute Frobenius and let $X' := X \times_{F_S} S$ be the Frobenius twist of $X$.
Question: are the sheaves $F^* \mathcal O_X$ and $\mathcal O_{X'}$ the same (as sheaves of $k$-algebras)?
Note that $X$ and $X'$ have the same underlying topologial space, hence we can compare those sheaves. It seems to me that both sheaves are just $\mathcal O_X$ with a "funny" $k$-algebra structure:
$$(\lambda, x) \mapsto \lambda^{1/p} \cdot x$$
-hence the question.
For an arbitrary scheme $S$ (not necessarily perfect) of characteristic $p$ this isn't true in general, right?