We say $X$ has property $(*)$ if $X$ is Nötherian, integral, separated, and regular in codimension one.
My question is the following: If $\imath:Y\to X$ and $\jmath:Z\to X$ are closed immersions and $Y$ and $Z$ are integral schemes such that there exists a homeomorphism $\lvert\varphi\rvert:\lvert Y\rvert\to \lvert Z\rvert$ such that $\lvert\jmath\rvert\circ\lvert\varphi\rvert=\lvert\imath\rvert$, then can $\lvert\varphi\rvert$ be extended to a morphism of schemes $\varphi:Y\to Z$ such that $\jmath\circ\varphi=\imath$?
My intuition for why this would be true is that the most common instance in which we have several scheme structures on a closed topological subset of $X$ is when we study non-reduced schemes, and I know of no examples of two distinct integral scheme structures on a closed topological subset of $X$. However, I don't know how anything like this would be proved.
The answer is yes - all you need is that the closed subscheme is reduced. Ie, you don't need any noetherian, separatedness, regularity, or irreducibility assumptions.
Closed subschemes of $X$ are in bijection with quasicoherent sheaves of ideals on $X$ (c.f. Hartshorne Prop II.5.9). If two quasicoherent sheaves of ideals $\mathcal{I},\mathcal{J}$ are not the same, then they must differ on some open affine $\text{Spec }A$, where they are just usual ideals $I,J$ of the coordinate ring $A$. Thus, it suffices to study the affine case.
For any ideal $I$ of a ring $A$, we may consider its radical $rad(I)$, see for example: wikipedia. The set of primes of $A$ which contain $I$ are the same as those which contain $rad(I)$, and $A/I$ is reduced if and only if $I = rad(I)$ (ie, $I$ is a radical ideal). This implies that there is a bijection between closed subspaces of $\text{Spec }A$ and radical ideals of $I$, each of which gives rise to a reduced closed subscheme, and all the nonradical ideals correspond to nonreduced closed subschemes, which have the same topological space as their radical.