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](https://mathoverflow.net/questions/354193/are-the-rings-mathbbcx-langle-x2-c-mathrmtrx-x-rangle-isomorphic/354225#354225). Indeed, the existence of such a lift would guarantee the isomorphism between the rings introduced by the OP in that other question. 



schemes having same reduced underlying space and same cotangent sheaf are isomorphic?
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 $X \not\cong Y$.


