Let $X$ be a normal scheme with quotient singularities and $Y\subset X$ its singular locus. The first order deformations of $X$ are parametrized by $\mathcal{E}xt^{1}(\Omega\_{X},\mathcal{O}\_{X})$. This sheaf is supported on $Y$.


I would like to relate $\mathcal{E}xt^{1}(\Omega\_{X},\mathcal{O}\_{X})$ and $\mathcal{E}xt^{1}(\Omega\_{Y},\mathcal{O}\_{Y})$ (here I am considering first order deformations of $Y$ as an abstract scheme not its embedded deformations as a subscheme of $X$). To do this I am thinking to consider the conormal exact sequence 
$$\mathcal{I}\_{Y}/\mathcal{I}\_{Y}^{2}\rightarrow \Omega\_{X|Y}\rightarrow\Omega\_{Y}\mapsto 0$$
Applying $\mathcal{H}om(-,\mathcal{O}\_{Y})$ we get
$$0\mapsto \mathcal{H}om(\Omega\_{Y},\mathcal{O}\_{Y})\rightarrow \mathcal{H}om(\Omega\_{X|Y},\mathcal{O}\_{Y})\rightarrow \mathcal{H}om(\mathcal{I}\_{Y}/\mathcal{I}\_{Y}^{2},\mathcal{O}\_{Y})\rightarrow \mathcal{E}xt^{1}(\Omega\_{Y},\mathcal{O}\_{Y})\rightarrow \mathcal{E}xt^{1}(\Omega\_{X|Y},\mathcal{O}\_{Y})\rightarrow \mathcal{E}xt^{1}(\mathcal{I}\_{Y}/\mathcal{I}\_{Y}^{2},\mathcal{O}\_{Y})\rightarrow ....$$
*Is there any reason why one should have $\mathcal{E}xt^{1}(\mathcal{I}\_{Y}/\mathcal{I}\_{Y}^{2},\mathcal{O}\_{Y}) = 0$ or $\mathcal{H}om(\mathcal{I}\_{Y}/\mathcal{I}\_{Y}^{2},\mathcal{O}\_{Y}) = 0$ ?*



Relating deformations of a scheme to deformations of its singular locus
Here is a partial answer. 


Let $Y$ be a smooth variety over a field $k$ of characteristic zero. Let $G$ be a finite group acting on $Y$, and let $X = Y/G$ be the quotient. Assume that the set of points where the isotropy is not trivial is in codimension greater or equal than three, that is the singular locus of $X$ is in codimension greater or equal than three. Then $Ext^1(\omega\_X,\mathcal{O}\_X) = 0$, that is $X$ is rigid.


One can find this in:


* M. Schlessinger, *"Rigidity of quotient singularities"*, Inventiones mathematicae, 1971, Volume 14, Issue 1, pp. 17-26.


This could fail is $X$ is singular in codimension two. For instance, we may consider the a singular point of type $\frac{1}{6}(2,4)$. Then, étale locally, in a neighborhood of $p$ the surface $X$ is isomorphic to $\mathbb{A}^{2}/\mu\_{6}$ where the action is given by 
$$
\begin{array}{ccc}
\mu\_{6}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\
(\epsilon,x\_{1},x\_{2}) & \longmapsto & (\epsilon^{2}x\_{1},\epsilon^{4}x\_{2})
\end{array}
$$
The invariant polynomials with respect to this action are clearly $x\_{1}^{3},x\_{2}^{3},x\_{1}x\_{2}$. Therefore, étale locally, in a neighborhood of $p$ the surface $X$ is isomorphic to an étale neighborhood of singularity
$$S = \{f(x,y,z) = z^{3}-xy = 0\}\subset\mathbb{A}^{3}.$$
Now, we have 
$$Ext^{1}(\Omega\_{S},\mathcal{O}\_{S})\cong K[x,y,z]/(f,\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}) = K[x,y,z]/(z^{3}-xy,-y,-x,3z^{2})\cong K[z]/(z^{2}).$$


Therefore $X$ is not rigid.


