schemes having same reduced underlying space and same cotangent sheaf are isomorphic?

Question

Let $X,Y$ be two closed subschemes of $\mathbb{A}^n_{\mathbb{C}}$ for some fixed $n$. Assume that I have an isomorphism of the underlying reduced spaces: $$f : X_\text{red} \stackrel{\sim}\longrightarrow Y_\text{red}$$ which induces an isomorphism of $\mathcal{O}_{X_\text{red}}$-module:

$$ f^{*} \left(\Omega_{Y/\mathbb{C}} \otimes \mathcal{O}_{Y_\text{red}} \right) \stackrel{\sim}\longrightarrow \Omega_{X/\mathbb{C}} \otimes \mathcal{O}_{X_\text{red}}.$$

Can I lift $f$ to an actual isomorphism $X \stackrel{\sim}\longrightarrow Y$? Note that in the situation I am interested $Y$ is very singular. If $Y$ were smooth, I guess the theory of infinitesimal liftings would give the answer.

This is related to this question. Indeed, the existence of such a lift would guarantee the isomorphism between the rings introduced by the OP in that other question.

Answer 1

Consider the simplest example: $$ X = \mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^2), \qquad Y = \mathrm{Spec}(\mathbb{C}[\epsilon]/\epsilon^3). $$ Definitely, $X_{\mathrm{red}} \cong Y_{\mathrm{red}}$. Also, a simple computation shows that $$ \Omega_{X/\mathbb{C}} \cong \mathbb{C}, \qquad \Omega_{Y/\mathbb{C}} \cong \mathbb{C}[\epsilon]/\epsilon^2, $$ hence $\Omega_{X/\mathbb{C}} \otimes \mathcal{O}_{X_{\mathrm{red}}} \cong \Omega_{Y/\mathbb{C}} \otimes \mathcal{O}_{Y_{\mathrm{red}}}$, but of course $X \not\cong Y$.