Solving a functional equation arising from a probability problem: $g(kx)^2=g(x)$ I am trying to find solutions to the following functional equation:
$$g(kx)^2=g(x)\text.$$
Here, $x$ is in $\mathbb{R}$ and $k$ is a constant. In particular, I'm looking for solutions for $k=2^{-1/4}$. Furthermore, I need that the Fourier inverse of $g$ is a density (i.e., nonnegative and integral over $\mathbb{R}$ is 1). The function $g(x)=e^{-x^4}$ satisfies the equation, but does not have a nonnegative Fourier inverse. I have deduced that $g(0)$ must be one, but have little experience with functional equations and am at a loss at how to proceed. (It may very well be the case that there are no other solutions.)
The context is the following. I am tasked with finding (or showing that there exist none) i.i.d. random variables $X$ and $Y$ such that $\frac{X+Y}{2^{1/4}}\sim X$. If you assume that $X$ and $Y$ have density $f$, then standard Fourier arguments show that $\hat{f}=g$ must satisfy the functional equation above. The requirement that the Fourier inverse of $g$ be nonnegative comes from the fact that $f$ is a density.
Any hint, either with the functional equation or the original problem, would be greatly appreciated. In particular, with regards to the original problem, I have been able to deduce that $E[X]$ is either 0 or infinite, and in either case, $E[X^2]$ is infinite, but I am not sure how to proceed to show either existence or non-existence.
 A: Unfinished Attempt: I shall return!
For a fixed $k\in\mathbb{R}$, I shall find the general solution $g:\mathbb{R}\to\mathbb{R}$ to the functional equation$$\big(g(kx)\big)^2=g(x)\text{ for all }x\in\mathbb{R}\,.\tag{*}$$
It can be easily seen that $g(0)\in\{0,1\}$.  If $k=0$, then it follows immediately that $g\equiv 0$ and $g\equiv 1$ are the only solutions.   
If $k=\pm1$, then we get $g(x)\in\{0,1\}$ for all $x\in\mathbb{R}$ (with the additional requirement that $g$ be an even function in the case $k=-1$).  From now on, assume that $k\notin\{-1,0,+1\}$.  Clearly, we have $g(x)\geq 0$ for every $x\in\mathbb{R}$.
If $k>1$, then $g$ is completely determined by its behavior on $[+1,+k)$ and $(-k,-1]$ using (*), as well as the exceptional value $\epsilon:=g(0)\in\{0,1\}$.  Let $g_+:[+1,+k)\to\mathbb{R}_{\geq 0}$ and $g_-:(-k,-1]\to\mathbb{R}_{\geq 0}$ be arbitrary.  We have, for each $x\in\mathbb{R}$,
$$g(x)=\begin{cases}\sqrt[2^{n(k,x)}]{g_+\left(\frac{x}{k^{n(k,x)}}\right)}&\text{if }x>0\,,
\\\epsilon&\text{if }x=0\,,\\
\sqrt[2^{n(k,x)}]{g_-\left(\frac{x}{k^{n(k,x)}}\right)}&\text{if }x<0\,.
\end{cases}\tag{#}$$
Here, $$n(k,x):=\left\lfloor\frac{\ln|x|}{\ln(k)}\right\rfloor\text{ for each }x\in\mathbb{R}\,.$$
If $0<k<1$, then $g$ is completely determined by its behavior on $(+k,+1]$ and $[-1,-k)$, as well as the value $\epsilon:=g(0)\in\{0,1\}$.  Let $g_+:(+k,+1]\to\mathbb{R}_{\geq 0}$ and $g_-:[-1,-k)\to\mathbb{R}_{\geq 0}$ be arbitrary.  Then, $g$ is also given by (#).
This leaves the case $k<0$ (with $k\neq -1$).  I still need to look into the Fourier transform condition.
