I've been reading a bit about Lebesgue measurability in real analysis. The common example of Vitali Sets came up to demonstrate why we can't have any old subset of $\mathbb{R}$ be Lebesgue measurable.
I've seen another example of non-measurable sets involving rational rotations of all the points about a circle.
To me, the set of points about the unit circle and the set of points on the $[0,1]$ seem equivalent, in which case, rational rotations are the same as the rational shifts of Vitali's theorem. In both cases, we have an uncountable set that that is partitioned into a countably infinite set via some quotient group (e.g., $\frac{\mathbb{R}}{\mathbb{Q}}$) (plus Axiom of Choice!), which leads us to having to conclude the measure is $0$ or $\infty$, neither of which equal the (known) Lebesgue measure of the set being partitioned.
Is there a theorem or sufficient/necessary set of conditions that can back this up?
EDIT: Per the comments below by @user85667 I think I need to be more precise when I say "isomorphic" or “similar to” Vitali Sets. Here is the particular sense I am thinking:
Conjecture
All non-Lebesgue-measurable sets on $\mathbb{R}$ are dense elements of an uncountable quotient group.
Best Answer
Your conjecture is currently far too imprecisely stated to be disproved, but certainly there are naturally-occurring types of nonmeasurable set (that is, the sets themselves might not be natural, but their key properties are) which are quite different from Vitali sets.
For example, there are Bernstein sets. A set $A$ is Bernstein iff both $A$ and $A^c$ (= the complement of $A$) have nonempty intersection with every perfect (= nonempty closed without isolated points) set. Bernstein sets need not be Vitali. There are also Hamel bases (= bases for $\mathbb{R}$ as a vector space over $\mathbb{Q}$); no Hamel basis can be Vitali, and non-null Hamel bases must be non-measurable (there do exist null, so a fortiori measurable, Hamel bases). And there are many more types of pathological set, such as Luzin sets and Sierpinski sets. See these notes of Beriashvili.
Basically, the sheer variety of interesting types of non-measurable sets out there strongly suggests to me that there is no sense in which all instances of non-measurability must be related to Vitali sets.