Suppose we have two (nondegenerate) probability distributions on $\mathbb{R}$, $P$ and $\widetilde P$, such that for $X, Y \stackrel{\text{iid}}{\sim} P$ and $\widetilde X, \widetilde Y \stackrel{\text{iid}}{\sim} \widetilde P$ $$X-Y \stackrel{d}{=} \widetilde X - \widetilde Y$$ where the equality is in distribution. What can I say about the relationship between the two distributions $P$ and $\widetilde P$? Of course, they are not necessarily equal, since if we fix a distribution $P$, $X \sim P$ and let $\widetilde P$ be the law of $X+c$ for an arbitrary constant, then we have two different distributions (although trivially different) which satisfy the requirement. Is this the only kind of difference possible?
For context: I am trying to give a rigorous proof of the general form of the characteristic function of a stable distribution, and I want to use this to go from the general form of a strictly stable ch.f. (which I have a rigorous proof of) to the general form of a non-strictly stable ch.f.
Let $\varphi$ denote the characteristic function of $P$ and $\widetilde{\varphi}$ that of $\widetilde P$. The characteristic function of $X-Y$ is $$ t\mapsto \varphi(t)\varphi(-t)=\left\lvert \varphi(t)\right\rvert^2 $$ and that of $X-Y$ is $$ t\mapsto \widetilde{\varphi}(t)\widetilde{\varphi}(-t)=\left\lvert \widetilde{\varphi}(t)\right\rvert^2. $$ Therefore, $X-Y \stackrel{d}{=} \widetilde X - \widetilde Y$ is equivalent to $$\forall t\in\mathbb R, \quad \left\lvert \varphi(t)\right\rvert=\left\lvert \widetilde{\varphi}(t)\right\rvert. $$