When is the product XY of two continuous random variables uncorrelated with X? Let $X$, $Y$ be two independent random variables defined on the real line by a fixed distribution $\mathcal{D}$. Let $Z = XY$, the product of $X$ and $Y$.
For which choices of $\mathcal{D}$ is $Z$ uncorrelated with the individual variables $X, Y$, so $\rho_{Z, X} = \frac{\text{cov}\{Z, X\} } {\sigma_{Z} \sigma_{X}}  = 0$?
As an example, for $X, Y \sim \mathcal{N}(0,1)$
> N <- 1000000
> set.seed(1)
> x <- rnorm(N)
> y <- rnorm(N)
> xy <- x*y
> cor(xy, x)
[1] -0.0001072026

this seems to be the case.
I would be grateful for pointers to (i) a derivation of this for the Gaussian case and (ii) any results regarding the characteristics of the density that determine whether this independence holds.
 A: For iid random variables $X$ and $Y$ with finite variance, set $Z = XY$, and note that $Z$ also has finite variance as can be deduced, for example, from the formulas in this question and its answers.  Consequently, $\rho(Z,X)$ equals $0$ if and only if $\operatorname{cov}(Z,X)$ equals $0$.  But,
\begin{align}
\operatorname{cov}(Z,X) &= E[ZX]-E[Z]E[X]\\
&= E[X^2Y]-E[XY]E[X]\\
&= E[X^2]E[Y]-E[X]E[Y]E[X] &\scriptstyle{X,Y}~\text{independent}\implies X^2,Y~\text{also independent}\\
&= \left(E[X^2]-(E[X])^2\right)E[Y]\\
&= \sigma_X^2E[Y]\\
&=0 ~~\text{if and only if }~\sigma_X^2 = 0 ~ \text{or} ~E[Y]=0.
\end{align}
Similarly, $\operatorname{cov}(Z,Y)=0$ if and only if $\sigma_Y^2 = 0$ or $E[X]=0$.  Of course, since $X$ and $Y$ are identically distributed, $\sigma_X^2=\sigma_Y^2$ and $E[X]=E[Y]$.
In short, what @whuber's comment calls a "(trivially) true" condition for the result to hold is not just sufficient but also necessary except in the special case when the common distribution has zero variance and so both $X$ and $Y$ almost surely equal the same constant (which is also their common expected value) and this constant (a.k.a. $E[X]=E]Y]$) need not be $0$.
