# 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 $$X$$?

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.