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. 
 A: There is$^*$ a counterexample in the atomic case, see below, so we will assume that $(\Omega, \mathrm{P})$ is a non-atomic standard Lebesgue probability space (so it is Isomorphic to the unit interval with Lebesgue measure, see https://en.wikipedia.org/wiki/Standard_probability_space). In that setting we claim:
If $X \overset{\mathrm{d}}{=} Y$ are two random variables on $(\Omega, \mathrm{P})$ , then there always exists a nondegenerate random variable $Z$ on $(\Omega, \mathrm{P})$ satisfying $X + Z \overset{\mathrm{d}}{=} Y + Z$.
Proof: We consider three cases. 
Case 1 : $X+Y$ is not an almost sure constant. Then defining $Z:=-X-Y$ proves the claim.
Case 2: $X+Y=c$ almost surely but $X$ takes more than 2 values a.s. (That is, for every pair of reals $a,b$ we have $  \mathrm{P}[X \in\{a,b\}]<1.$ ) Then $Z:=X(c-X)$ is nondegenerate. Denoting   $f(x)= x+x(c-x)$, we have
$f(X)=X+Z$ and $f(Y)=Y+Z$ almost surely, so $X + Z \overset{\mathrm{d}}{=} Y + Z$.
Case 3: $X+Y=c$ almost surely, and there exist $a,b$ such that $  \mathrm{P}[X \in\{a,b\}]=1.$ In this case let $D_a$ be a subset of $\{\omega: X(\omega)=a\}$ with 
$\mathrm{P}(D_a)=\mathrm{P}(\{\omega: X(\omega)=a\})/2$. (To see that $D_a$ exists we may assume that $(\Omega, \mathrm{P})$ is the unit interval with Lebesgue measure, and observe that the mapping $t \mapsto \mathrm{P}([0,t] \cap \{\omega: X(\omega)=a\})$ is continuous on $[0,1]$ so its image includes $\mathrm{P}(\{\omega: X(\omega)=a\})/2$.) Similarly, let $D_b$ be a subset of $\{\omega: X(\omega)=b\}$ with 
$\mathrm{P}(D_b)=\mathrm{P}(\{\omega: X(\omega)=b\})/2$. Let $Z(\omega)=1$ if $\omega \in D_a \cup D_b$ and $Z(\omega)=0$ otherwise. Then $Z$ is independent of $X,Y$ so $X + Z \overset{\mathrm{d}}{=} Y + Z$.
QED
(*) Counterexample in the atomic case: 
If $\Omega=\{0,1\}$ with the uniform measure and maximal $\sigma$-algebra, and $X(\omega)=\omega=1-Y(\omega)$, then for every nonconstant random variable $Z$ on $\Omega$, the laws of $X+Z$ and $Y+Z$ are different; e.g., if $Z(1)>Z(0)$ then the maximum of $X+Z$ exceeds the maximum of $Y+Z$.
