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 $\eta\neq\mathrm{id}_X$ 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 $\mathrm{id}_X$ 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, solutions of $(1)$ which are bijection form a group - which may be thought of a group of "symmetries" of $\mathsf Q$, apparently.
A small example just to add some clarity to the problem statement.
If $X = \{a,b\}$, $\mathscr A = 2^X$ and $\mathsf Q(a) = 0.4$ then the solution is unique. However, if $\mathsf Q(a) = 0.5$ there are exactly two solutions. Indeed, there are exactly $4$ maps $\xi:X\to X$ namely $$ \begin{align} \xi^1:(a,b) \mapsto (a,a) & &\xi^2:(a,b) \mapsto (a,b) \\ \xi^3:(a,b) \mapsto (b,a) & &\xi^4:(a,b) \mapsto (b,b) \end{align} $$ In the first case, the pushforwards are $$ \begin{align} \xi^1_*(\mathsf Q) &= &(1,0) & &\xi^2_*(\mathsf Q) &= &(0.4,0.6) \\ \xi^3_*(\mathsf Q) &= &(0.6,0.4) & &\xi^4_*(\mathsf Q) &= &(0,1) \end{align} $$ and in the second case: $$ \begin{align} \xi^1_*(\mathsf Q) &= &(1,0) & &\xi^2_*(\mathsf Q) &= &(0.5,0.5) \\ \xi^3_*(\mathsf Q) &= &(0.5,0.5) & &\xi^4_*(\mathsf Q) &= &(0,1) \end{align} $$ hence in the first case $(1)$ has the only solution $\xi^2$ which is of course $\mathrm{id}_X$, but in the second case both $\xi^2$ and $\xi^3$ solve the problem.
Due to this reason, I expect a non-uniqueness of the solution to reflect some kind of symmetries in the distribution $\mathsf Q$.
I posted the same question on MO.
