Finding all Borel measures $\mu_X$ such that $Y\sim \mathcal{N}(0,1) \Rightarrow XY \sim \mathcal{N}(0,1)$. 
Find all Borel measures $\mu$ on $\mathbb{R}$ such that for every independent random variables such that $X \sim \mu$ and $Y\sim \mathcal{N}(0,1)$ we have $XY \sim \mathcal{N}(0,1)$.

To be honest I have no idea how to bite this problem so every help will be appreciated.
$\mathcal{N}(0, 1)$ means distribution with $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ as density.
 A: If $\mu$ has a finite fourth moment, use $E((XY)^2)=1=E(Y^2)$ and $E((XY)^4)=3=E(Y^4)$, hence $E(X^2)=E(X^4)=1$. This implies that $E((X^2-E(X^2))^2)=0$, thus $X^2=E(X^2)$ is almost surely constant. Since $E(X^2)=1$, $X=\pm1$ almost surely. 
Conversely, every Bernoulli distribution $\mu=p\delta_1+(1-p)\delta_{-1}$ is a solution.
Otherwise, use $E(\mathrm e^{\mathrm itXY})=\mathrm e^{-t^2/2}$ and the identity $E(\mathrm e^{\mathrm itXY})=E(E(\mathrm e^{\mathrm itXY}\mid X))=E(\mathrm e^{-t^2X^2/2})$ for every real number $t$. This shows that the Laplace transform of $X^2$ is $E(\mathrm e^{-\lambda X^2})=\mathrm e^{-\lambda}$ for every $\lambda\geqslant0$, thus, $X^2=1$ almost surely, leading to the same conclusion.
A: Notice that $XY$ has a finite fourth moment order, and by independence, so does $X$. Denote by $m_k:=\mathbb E[Z^k]$ where $Z$ is a standard normal random variable. Then 
$m_2=\mathbb E[X^2Y^2]=\mathbb E[X^2]\cdot m_2$ and since $m_2\neq 0$, $\mathbb E[X^2]=1$. Similarly, we get that $\mathbb E[X^4]=1$ and this implies that $X^2$ is almost surely constant, say to $c^2$. Notice that $c=\pm 1$. It remains to determine the probabilities.
