Let $k$ be an algebraically closed field (not necessarily of characteristic $0$), $X$ a non-singular affine closed subscheme in $\mathbb{A}^n\_k$ for some $n \ge 2$. Denote by $I\_X$ the ideal of $X$ in $\mathbb{A}^n\_k$. Let $X'$ be a first order infinitesimal deformation of $X$ i.e., $X'$ is flat over $\mathrm{Spec}(k[t]/(t^2))$ having $X$ as the closed fiber. Denote by $I\_{X'}$ the ideal of $X'$ in $\mathbb{A}^n\_k \times \mbox{Spec}(k[t]/(t^2))$. By the infinitesimal lifting property, we know that $X'$ is isomorphic to $X \times\_k \mbox{Spec}(k[t]/(t^2))$. But this isomorphism is not canonical. The question is whether there is an isomorphism between $X \times\_k \mbox{Spec}(k[t]/(t^2))$ and $X'$ such that the induced morphism of ring $$\frac{k[X\_1,...,X\_n]}{I\_X} \otimes\_k k[t]/(t^2) \to \frac{(k[t]/(t^2))[X\_1,...,X\_n]}{I\_{X'}}$$ sending $X\_i$ to $X\_i$? Furthermore, can we get such an isomorphism for which $t$ maps to $t$ as well? 



On the infinitesimal lifting property of non-singular affine schemes
The answer is no. Easiest nontrivial case: $n=1$, $I\_X=(X\_1)$, so $X=Spec(k[X\_1]/(X\_1))=Spec(k)$, take $I\_{X'} = (X\_1 - t)$, so that $X' = Spec(k[X\_1, t]/(X\_1 - t, t^2))$. Now you are asking for an isomorphism $k[X\_1, t]/(X\_1, t^2)\to k[X\_1, t]/(X\_1 - t, t^2)$ sending $X\_1$ to $X\_1$, but this cannot exist as $X\_1=0$ on the left while $X\_1\neq 0$ on the right.


