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Yes. Choose the principal ultrafilter at any $j\ge 2$. Evaluation at the $j$th coordinate giveswhich is a finite cyclic group of order $j$.

The scalar sign action of a cyclic group of order two on ${\mathbb{C}}^2$ has invariant ring $\mathbb{C}[X^2,XY,Y^2]\cong \mathbb{C}[a,b,c]/(ac-b^2)$. It is not factorial: $a,b,c$ are pairwise nonassociate irreducibles, since every nonconstant invariant has degree at least two, while $ac=b^2$ gives two distinct factorizations.