Are the r.v. such that their sum produces the same distribution Suppose, $X_1,X_2$ are i.i.d random variables with some distribution $F_X(x)$.
Let $Y=X_1+X_2$. Does there exist $F_X(x)$ such that $F_X(x)=F_Y(x)$ or maybe $F_X(x)=F_Y(x)$ a.e.?
Thanks. Sorry, for the vague question I am still trying to formalize it.I will be happy to answer any of your questions. 
 A: If $X_1$ is square integrable, then we have $EY^2 = 2EX_1^2$. If $X_1$ and $Y$ have the same distribution function, they have the same second moment, so the only possibility is $EX_1^2 = 0$, i.e. $X_1 = 0$ almost surely.
If $X_1$ is not square integrable but $E|X|<+\infty$, from $EY = E(X_1 + X_2) = EX_1$, we get $EX_1=EX_2 = 0$, then we have $E(Y|X_1) = X_1$, which means the conditional expectation of $Y$ has the same distribution with $Y$, then from this question, we have $Y= X_1$ almost surely, i.e. $X_2 = 0$ almost surely.
So $X_1 = X_2 = 0$ is the only possibility for $L^1$ variable
A: Not without any renormalization.
Indeed, if for instance
$$
E[X^2]<\infty
$$then you should have
$$
\text{var }X = \text{var }X_1 + \text{var }X_2= 2\text{var }X
$$leading to $\text{var }X = 0$.

if you allow an identity such as
$$
X_1 + X_2 = aX
$$then there are several example of such distributions:
the Normal distribution, and the Cauchy distribution.

For more information on this subject, you can take a look at the wikipedia page about stable distributions.
A: The result holds without integrability condition: 
Consider the characteristic function $\varphi$ of $X$, defined for every real number $x$ by $\varphi(x)=E(\mathrm e^{\mathrm ixX})$, then the hypothesis implies the identity $$\varphi(x)=E(\mathrm e^{\mathrm ix(X_1+X_2)})=E(\mathrm e^{\mathrm ixX_1})E(\mathrm e^{\mathrm ixX_2})=\varphi(x)^2,$$ for every $x$, hence $\varphi(x)\in\{0,1\}$. One knows that $\varphi(0)=1$ and that $\varphi$ is continuous hence $\varphi(x)=1$ for every $x$. As the name suggests, $\varphi$ characterizes the distribution of $X$ and the Dirac mass at zero has this characteristic function hence $X=0$ almost surely.
