# Sum of random variables are equal in distribution

Suppose that $$X,Y$$ are scalar random variables supported on some standard Lebesgue probability space $$(\Omega, \mathrm{P})$$, such that $$X \overset{\mathrm{d}}{=} Y$$ in the sense that their pushforward measures are equal, $$X_*(\mathrm{P}) = Y_*(\mathrm{P})$$. Does there exist a nondegenerate random variable $$Z$$ on $$(\Omega, \mathrm{P})$$ satisfying $$X + Z \overset{\mathrm{d}}{=} Y + Z$$?

In the case that $$Y = X\circ T$$ for some measure preserving automorphism $$T: \Omega \rightarrow \Omega$$ (this appears to be often the case by: Random variables with same distribution), and we have in addition the factorization $$T = S^2$$ for some automorphism $$S: \Omega \rightarrow \Omega$$, then clearly we can set $$Z = X \circ S$$, whence

$$X + Z = X + X\circ S \overset{\mathrm{d}}{=} X \circ S + X \circ S^2 = Z + Y$$.

Are there more general conditions than this, or perhaps conditions under which the answer to the question is negative?

Edit: it also turns out that if $$Y = X \circ T$$ where $$T$$ is measure preserving and non-ergodic then $$Z$$ exists; just take $$Z$$ such that $$Z \circ T = Z$$ almost surely and $$Z$$ non-constant. Then a computation like the one above shows equality in distribution.