Let me define a closed subscheme of a scheme $X$ as:
An equivalence class of data in the form
$$(Z,Y,i,i^\sharp)$$
where $Z$ is a closed subset of $X$, $(i,i^\sharp): Y \to X$ is a morphism of schemes with $i$ giving a homeomorphism onto $Z$, and $i^\sharp : \mathcal{O}_X \rightarrow i_* \mathcal{O}_Y$ is surjective.
We say $(Z_1,Y_1,i_1,i_1^\sharp)$ and $(Z_2,Y_2,i_2,i_2^\sharp)$ are equivalent if there is an isomorphism of schemes $\rho$ from $Y_1$ to $Y_2$ such that $(i_1,i_1^\sharp)=\rho \circ (i_2,i_2^\sharp)$. Clearly, these two data are equivalent only if $Z_1$ and $Z_2$ are the same subset of $X$.
Can we have another closed subscheme of $X$ on the same exact closed subset $Z$, say $(Z,Y,j,j^\sharp)$, that is not equivalent to $(Z,Y,i,i^\sharp)$, with the same source scheme?
If yes, then is it possible to have an example where $j=i$ actually?
Thanks.
You seem to ask if there are two closed subschemes $Y$, $Y'$ of a scheme $X$ which are not isomorphic over $X$, but $Y \cong Y'$ as abstract schemes and $|Y|=|Y'|$ as subsets of $|X|$.
If $X=\mathrm{Spec}(A)$ is affine, this means that we have to find two ideals $I,J \subseteq A$ such that $I \neq J$, but $A/I \cong A/J$ as rings and $\sqrt{I} = \sqrt{J}$.
It suffices to find some ideal $I$ with $A/I \cong A/I^2$ and $I \neq I^2$.
Take $A=k[X_1,X_2,\dotsc]/(X_1^2,X_2^2,\dotsc)$ and $I=(X_1)$. Then $I^2 = 0 \neq I$, and $A/I = k[X_2,\dotsc]/(X_2^2,\dotsc) \cong A$.
I wonder if there is any example for $A$ noetherian.