# Is there a “minimal” center of a blowup?

Let $$X$$ be a scheme, let $$i : Z \to X$$ be a closed subscheme, let $$Y := \mathrm{Bl}_{Z}(X)$$ be the blowup of $$X$$ at $$Z$$ with projection $$\pi : Y \to X$$. Suppose $$U \supseteq X \setminus Z$$ is an open subscheme of $$X$$ for which $$\pi^{-1}(U) \to U$$ is an isomorphism. Does there exist a closed subscheme $$Z' \to X$$ with support contained in $$X \setminus U$$ and for which the blowup $$Y' := \mathrm{Bl}_{Z'}(X)$$ is $$X$$-isomorphic to $$Y$$?

Motivation: If we blow up a subscheme of a normal scheme $$X$$, the maximal open subscheme $$U$$ for which $$\pi^{-1}(U) \to U$$ is an isomorphism contains all the codimension 1 points of $$X$$, so I was naively wondering whether all blow ups of a normal scheme are obtained by blowing up a closed subscheme of codimension at least 2.

Thoughts: If $$A$$ is a ring and $$I$$ is an ideal of $$A$$ and $$a \in A$$ is a nonzerodivisor, then the Rees algebras $$\bigoplus_{n \ge 0} I^{n}$$ and $$\bigoplus_{n \ge 0} (aI)^{n}$$ are isomorphic so if the center of the blowup is of the form $$\operatorname{Spec} A/(aI)$$, I can replace it with $$\operatorname{Spec} A/I$$, but I don't know how to recognize an ideal as being of the form $$aI$$.

Note that if $$Y = Bl_I(X)$$ then $$I$$, up to a twist and raising to a power, is the pushforward of a relative ample line bundle for $$Y \to X$$, so if the relative Picard number is 1, there are not so many choices for $$I$$.
Now, as an example, consider $$X = C(\mathbb{P}^1 \times \mathbb{P}^1)$$ be the cone over a smooth 2-dimensional quadric. Let $$Y = Tot_{\mathbb{P}^1}(\mathcal{O} \oplus \mathcal{O}(-1) \oplus \mathcal{O}(-1))$$ be one of its two small resolutions. Note that $$Y$$ is the blowup of a Weil divisor on $$X$$ --- the strict transform of one of the rulings of $$\mathbb{P}^1 \times \mathbb{P}^1$$. In particular, its ideal is NOT supported at the vertex of the cone (while the vertex is the complement of $$U$$ in this case). On the other hand, the relative Picard number is 1, hence any ideal $$I$$ such that $$Y \cong BL_I(X)$$ is equal to the ideal of the above Weil divisor up to a twist and a power. In particular, any such ideal is not supported at the vertex.