Let $X$ be a scheme. Does the forgetful functor

$$\mathrm{EtSch}/X \to \mathrm{Sch}/X$$

have a right adjoint $\tilde{Z} \leftarrow Z$? One might call $\tilde{Z}$ the étalification of $Z$.

If $X$ is the spectrum of a field $k$, the answer is yes, using 1) the equivalence of sites between étale $k$-schemes and continuous $G$-sets, where $G$ is the absolute Galois group of $k$, 2) the fact that every sheaf on $G$-sets is representable. Explicitly: If $Z$ is a $k$-scheme, let us denote by $Z_{\mathrm{sep}}$ the subset of all points $z \in Z$ such that $k(z)/k$ is finite and separable. Then $$\tilde{Z} = \coprod\limits_{z \in Z_{\mathrm{sep}}} \mathrm{Spec}(k(z)).$$ For example, $\widetilde{\mathbb{A}^1}=\coprod_{\alpha \in k} \mathrm{Spec}(k)$ and $\widetilde{\mathbb{A}^2}$ is the coproduct of spectra of the form $(k[x]/(p)[T])/(q)$, where $p \in k[x]$ is irreducible and $q$ is some irreducible separable polynomial over $k[x]/(p)$.

Perhaps this construction is well-known, therefore I've put the reference request tag.