= Solution
The <Absolute Frobenius morphism> is the identity on the underlying <topological space> and acts on the <structure sheaf> by $a\mapsto a^p$. On an <affine scheme> it is induced by the <ring homomorphism> $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a <prime ideal> $\mathfrak p$ is again $\mathfrak p$, since $a^p\in\mathfrak p$ exactly when $a\in\mathfrak p$. These affine maps agree under <localization> and hence glue.
A <perfect ring of characteristic p> is one whose <Frobenius endomorphism> is bijective. Since the <Absolute Frobenius morphism> fixes points, its map on the <stalk> at $x$ is precisely the <Frobenius endomorphism> of $\mathcal O_{X,x}$. A morphism of sheaves is an <isomorphism> exactly when it is an <isomorphism> on all <stalks>. Therefore
$$
\boxed{F_X\text{ is an isomorphism}\iff \mathcal O_{X,x}\text{ is perfect for every }x.}
$$
In the reverse direction the inverse <sheaf> map, together with the identity on points, gives the inverse <morphism of locally ringed spaces>. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.
For the <function field> assertions below, interpret the <normal schemes> as integral, as is implicit in writing a single <field> $k(X)$ or $k(Y)$. General <normal schemes> need not be irreducible; without this interpretation those <fields> are not defined globally as single <fields>. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\operatorname{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\operatorname{Spec}B$, with $A\subseteq B$ finite and both <rings> normal domains. In fact $B$ is the <integral closure> of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of <Frobenius factorization through a finite normal cover>.
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