What am I missing in the following:
Let $X=\mathbb{A}^2_k$ be the affine plane over an algebraically closed field $k$, and let $O$ be the origin. Let $\tilde{X}$ be the blow-up of $O$. If $\mathcal{I}$ is the ideal sheaf of $Y=\{O\}$ (with reduced induced structure), then the inverse image ideal sheaf $\mathcal{J}$ of $\mathcal{I}$ under the closed immersion $f:Y\to X$ is $0$. Hence the blow-up of $Y$ with respect to $\mathcal{J}$ is just $Y$ itself, i.e. the strict transform $\tilde{Y}$ is just $\tilde{Y}=Y$. But aren't there then plenty of morphisms $\tilde{f}:\tilde{Y}\to\tilde{X}$ such that $\pi_X\circ\tilde{f}=f\circ\pi_Y$, where $\pi_X:\tilde{X}\to X$ resp. $\pi_Y:\tilde{Y}\to Y$ are the natural maps? Because we can just send $\tilde{Y}$ to any point on the exceptional curve $E\subseteq \tilde{X}$ over $O$.
But then, this would contradict the unicity in Corollary II.7.15 of Hartshorne, so I'm sure I'm going wrong somewhere. But where?
As you correctly pointed out, the inverse image ideal sheaf $J$ is zero. But this means $$\tilde Y = \operatorname{Proj} \bigoplus_{d=0}^\infty J^d = \operatorname{Proj}(k \oplus 0 \oplus 0 \oplus \dots )= \emptyset$$ because the only homogeneous ideal in $k$ is the negligible one.