TITLE: Symmetries of probability distributions
QUESTION [9 upvotes]: When talking about a single random variable, knowing only its distribution, the construction of a probability space is quite easy. Namely, let $(X,\mathscr A)$ be a measurable space and let $\mathsf Q$ be some probability measure over this space which we refer to as a distribution of some random variable. The usual definition states that there is some probability space $(\Omega,\mathscr F,\mathsf P)$, the random variable is
$$
  \xi:(\Omega,\mathscr F)\to(X,\mathscr A)
$$
i.e. it is a measurable map, and its distribution is a pushforward measure: 
$$
\mathsf Q:=\xi_*(\mathsf P)
$$
i.e. $\mathsf Q(A) = \mathsf P(\xi^{-1}(A))$ for any $A\in \mathscr A$.
Clearly, given $(X,\mathscr A,\mathsf Q)$ for a single random variable there is no reason to come up with a new sample space and we can take $(\Omega,\mathscr F,\mathsf P) = (X,\mathscr A,\mathsf Q)$ and $\xi:=\mathrm{id}_X$. 
Let us stick to this latter case. It may happen, that there is a map 
$$
  \eta:(X,\mathscr A)\to(X,\mathscr A)
$$
such that $\eta\neq \rm id_X$ but still it holds that $\mathsf Q = \eta_*(\mathsf Q)$. I wonder if the existence of this other maps is studied somewhere.

The brief statement of the problem is thus the following: given a probability space $(X,\mathscr A,\mathsf Q)$ if the identity map $\rm id_X$ is the unique solution of the equation
$$
  \mathsf Q = \xi_*(\mathsf Q) \tag{1}
$$
where the variable $\xi$ is any measurable map from $(X,\mathscr A)$ to itself. As far as I am not mistaken, the space of solutions of $(1)$ is a monoid as it is closed under the composition of maps.
Also, if $\xi$ is a bijection which solves $(1)$ then $\xi^{-1}$ solves it as well:
$$
\xi^{-1}_*(\mathsf Q)(A) = \mathsf Q(\xi(A)) = \mathsf Q(\xi^{-1}(\xi(A))) = \mathsf Q(A).
$$
Hence, bijective solutions of $(1)$ form a group - which may seem to be thought of a group of "symmetries" of $\mathsf Q$. For example, the standard normal distribution over reals $\mathsf Q = \mathscr N(0,1)$ admits at least two representations $\xi(\omega) = \omega$ and $\xi(\omega) = -\omega$. As well as any Haar measure over a group admits representation via $\xi(\omega) = \alpha \omega$ where $\alpha$ is any element of the group.
I've asked this question on MSE, but I have not received any answers.

Edited: To clarify (as requested), my question is exactly as follows: are such groups of symmetries of measures studied somewhere in the literature - may be, providing some interesting results for measures exhibiting such symmetries. I have studied the Lie groups of ODE/PDE symmetries, and I wonder if there is anything similar known for measures.

REPLY [5 votes]: Usually, such symmetries have been either studied in the context of Lebesgue spaces or studied in the context of homogenous measure algebras, where autmorphisms are easy to study. Every automorphism of a probability space gives rise to an automorphism of the corresponding measure algebra.
The easiest case are Lebesgue spaces, or even simpler, studying the uniform distribution on $[0,1]$. This is of course essentially the case of a Haar measure. A nice property is that one can take any automorphism of the measure algebra and find an automorphism of the probability space inducing it. Moreover, two automorphisms of the probability space giving rise to the same automorphism of the measure algebra can differ only on a set of measure zero. Every, homogenous, atomless, separable measure algebra can be represented as a Lebesgue space.
If one starts with a homogenous measure algebra, one may look for probability spaces representing the measure algebra. The two most prominent representations are by the Stone space of the measure algebra or in the form of a product of coinflips $\{0,1\}^\kappa$ with $\kappa$ infinite, which can represent every atomless (normed) homogenous measure algebra by Maharam's theorem. In the case of the Stone space, the automorphisms of the measure algebra correspond essentially to the automorphisms of the representing probability space. In the coin-flipping representation, every automorphism of the measure algebra is induced by an automorphism of the probability space. But very different automorphisms may give rise to the same automorphism of the measure algebra. Actually, there exists an automorphism of $\{0,1\}^\mathfrak{c}$ that induces the identity on the measure algebra but has no fixed point.
The discussion so far is largely adapted from the introduction to Ergodic theory on homogeneous measure algebras. by Choksi and Prasad. The book this has appeared in is likely to be available somewhere on the internet...
One can also study the case of rigid probability spaces, where there is essentially no automorphism but the identity. It is actually possible to find a countably generated and countably separated measurable space in which all automorphisms differ from the identity only on a countable set. This is done in the booklet Borel Spaces by Rao and Rao in Proposition 4. There also is an example of an atomless, countably generated probability space with no autmorphism but the identity (up to a countable set) in Section 48 of Values of non-atomic games by Aumann and Shapley.