Relative Frobenius and absolute Frobenius Let $X$ is a scheme over a scheme $S$ of characteristic $p$. Let $F: X\to X^{(p)}$ be the relative Frobenius and let $W: X^{(p)}\to X$ be the natural map in fibre product. We know $W\circ F$ is the absolute Frobenius on $X$ by definition. What about $F\circ W$? Is is the absolute Frobenius on $X^{(p)}$ and why?
 A: sorry for joining the party so late! yes this is quite tricky to write down, but follows from the universal property of the relative Frobenius $F_{X}$. For any morphism $f: X\rightarrow Y$ over $S$ we have a commutative square:
$\require{AMScd}$
\begin{CD}
X^{} @>F_X>> X^{(p)}\\
@Vf VV @Vf^{(p)} VV\\
Y^{} @>F_Y>> Y^{(p)}
\end{CD}
For a proof see here, basically follows from definition and the fact that any ring hom $\phi$ is somehow "Frobenius equivariant", i.e. $\phi(a^p)=\phi(a)^p.$
Now replacing $X$ by $X^{(p)}$and $X^{(p)}$ by $X^{(p^2)}$ you get the diagram (you can do that because $X^{(p)},X^{(p^2)}$ are still by def. schemes over $S$).
\begin{CD}
X^{(p)} @>F_X^{(p)}>> X^{(p^2)}\\
@Vf VV @Vf^{(p)} VV\\
Y^{} @>F_Y>> Y^{(p)}
\end{CD}
Now in the lower row set $Y=X$ and $f=W$ following your definitions. Then you get
\begin{CD}
X^{(p)} @>F_X^{(p)}>> X^{(p^2)}\\
@VW VV @VW^{(p)} VV\\
X^{} @>F_X>> X^{(p)}
\end{CD}
Now observe $W^{(p)}F_X^{(p)}$ is the absolute Frobenius on $X^{(p)}$ by definition. By commutativity of the diagram, this is the same as $F_XW$
