If the variance of $X^2$ is zero, when is the variance of $X$ not zero $X$ is a random variable with mean $\mu$, variance $\sigma^2$, skewness $\gamma$ and kurtosis $\kappa$, and
$ \mathtt{Var}[X^2]=0.$
Under what conditions can $\mathtt{Var}[X]=\sigma^2 >0$?
Is my reasoning below correct or can it be done in a simpler way?
We have that
$$ 
\mathtt{Var}[X^2] = \sigma^2 \left( (\kappa-1)\sigma^2 + 4\mu \gamma \sigma + 4\mu^2                 \right ) =0.
$$
So $\sigma=0$ or
$$ \sigma = \frac{-2\mu \left [ \gamma \pm \sqrt{\gamma^2-(\kappa-1)} \right ]  }{\kappa-1}.$$
As $\sigma$ has to be real and non-negative, $\gamma^2 \ge \kappa-1$ (by definiton $\kappa-1>0$).
For example, if $\mu>0$ then we require negative skew, $\gamma<0$, to get a positve real solution for $\sigma$.

If $Var(X^2) = 0$ then $X^2 = a^2 = \text{constant}$. Now if $Var(X)>0$ then $X$ cannot be constant, but does take values given by $\pm a$ (with $a>0$).
This means that the only class of suitable distributions for $X$ are discrete and characterised by
\begin{equation}
X = \begin{cases}
    a, \text{with probability } p\\
    -a, \text{with probability} 1-p
    \end{cases}
\end{equation}
where $p \in (0,1)$.
You can find the mean, variance and other moments from this form.
