Dependence of strict transform on the subscheme along which we blow up It is stated in https://stacks.math.columbia.edu/tag/080C that

Note that taking the strict transform along a blowup depends on the closed subscheme used for the blowup (and not just on the morphism $S' \to S$).

I am looking for a positive result in this direction. I cannot think of any examples where the strict transform would depend on the chosen subscheme. I can only guess that such examples would be very strange or pathological.
If the scheme $S$ is a variety, does the strict transform still depend on the chosen closed subscheme?
More precisely: By "variety" we mean an integral separated scheme of finite type over an algebraically closed field. Let $S$ be a variety. Let $Z$ and $Z'$ be subschemes of $S$ such that the blowups $Bl_Z S \to S$ and $Bl_{Z'} S \to S$ are isomorphic over $S$. Let $X$ be a be a closed subscheme of $S$. Does the isomorphism $Bl_Z S \to Bl_{Z'} S$ over $S$ induce an isomorphism of the strict transforms of $X$?
 A: Yes, the strict transform still depends on the chosen subscheme even if $S$ is a variety. No, $\operatorname{Bl}_Z S \to \operatorname{Bl}_{Z'} S$ over $S$ does not induce an isomorphism of the strict transforms.
The strict transform depends on the chosen subscheme even in very tame situations. Remember that the strict transform of $X$ with respect to the blowup of $S$ along $Z$ is defined to be the (scheme-theoretic) closure of the inverse image of $X \setminus Z$. The blowup is necessarily an isomorphism on $S \setminus Z$, but it may happen that it is an isomorphism over a bigger open subset of $S$. The reason for defining the strict transform in this way is that then the strict transform is equal to the blowup of $X$ along $X \cap Z$.
Example 1. Let $S$ be the affine plane, $Z$ a closed integral curve and $Z'$ any other closed integral curve. The strict transform of $Z$ with respect to $\operatorname{Bl}_Z S \to S$ (the blowup of $S$ along $Z$) is empty, but the strict transform of $Z$ with respect to $\operatorname{Bl}_{Z'} S \to S$ is an integral curve. Note that $\operatorname{Bl}_Z S$ and $\operatorname{Bl}_{Z'} S$ are isomorphic over $S$, namely, they are both isomorphic to $S$.
Example 2. Let $X$ any closed subvariety of an affine space $\mathbb{A}^n := \operatorname{Spec} \mathbb{C}[x_1, \ldots, x_n]$ such that $X$ is not equal to the whole space. Let $V(f)$ any closed hypersurface of $\mathbb{A}^n$ containing $X$. Let $I$ any ideal such that topologically the intersection $V(I) \cap X_{\mathrm{top}}$ is not equal to $X_{\mathrm{top}}$. Then the strict transform $X$ with respect to the blowup of $S$ along $I$ is non-empty, while the strict transform of $X$ with respect to the blowup of $S$ along $f \cdot I$ is empty.
If $X$ is not integral, then depending on the chosen subscheme $Z$, the strict transform can have a different number of irreducible components or some embedded components might disappear.
What you probably want to consider instead is the scheme-theoretic closure of the inverse image of $X \setminus Y$, where $Y$ is the closed subset of $X$ where $S' \to S$ is not an isomorphism. This does not depend of the chosen subscheme $Z$ used to define the blowup. I hope this helps.
