Let $X$ be a connected, smooth affine algebraic variety over an algebraically closed field $K$ of characteristic zero. Assume we have a finite group $G$ acting on $X$ by morphisms of $K$-schemes. Choose a closed point $p\in X$. Is there a $G$-invariant affine open $U$, such that $p\in U$ and $\Omega\_{X/K}(U)$ is free over $O\_{X}(U)$? Notice that, in this case, the quotient space $X/G$ is also an affine $K$-variety, and that the quotient map $\pi: X\rightarrow X/G$ is finite. Furthermore, the quotient $\pi$ is étale at a point $p\in X$ if and only if $p$ is not a fixed point of any element $g\in G$. As this locus is open in $X$, my statement holds for every point $p$ in a dense open in $X$. If $p$ is fixed by every element in $G$, then choose an affine open $U$ containing $p$, which trivializes the sheaf of Kahler differentials. Then the open $V=\cap\_{g\in G}g(U)$ contains $p$ and satisfies the property. I have not been able to show that this holds for points that are fixed by a proper, non-trivial subgroup of $G$.


Notice that this statement is equivalent to showing that $\pi\_{\*}\Omega\_{X}$ is a finite locally free sheaf over $\pi\_{\*}O\_{X}$.



On the trivialization of the sheaf of kahler differentials on the G-invariant topology
Just to make the discussion in the comments a little more concrete: given the orbit $A=G\cdot p$, we can choose an affine open containing these points, and find functions $f\_a$ for $a\in A$ on this open such that $f\_a(a')=\delta\_{a,a'}$ for $a,a'\in A$ (by [Sun-tzu's remainder theorem](https://en.wikipedia.org/wiki/Chinese_remainder_theorem)). For any locally free sheaf $\mathcal{F}$, we can choose a set of sections $\sigma\_{1,a},\dots, \sigma\_{n,a}$ on $U$ that give a basis of the residue $\mathcal{F}\_a$. The sums $\sum\_{a\in A}f\_a\sigma\_{k,a}$ are then sections that give a basis of the fiber on a possibly smaller open subset $U'$ containing $A$. If you're careful about how you choose everything with respect to $G$, I believe you can get that $U'$ is manifestly $G$-invariant, but you can also just intersect it with all its images under $G$.


