Variance of $\bar S^2$ Let $X_1, ... , X_n \stackrel{i.i.d} \sim \operatorname{Normal}(\mu , \sigma^2)$. Show that $$\mathrm{Var}\left(\frac1n \sum_\limits{i=1}^n (X_i - \overline X)^2 \right) = \cfrac{2(n-1)\sigma^4}{n^2}$$
My attempt:
$$\mathrm{Var}\left(\frac1n \sum_\limits{i=1}^n (X_i - \overline X)^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n (X_i - \overline X)^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n X_i^2 - 2\overline X\sum\limits_{i=1}^n X_i + \sum\limits_{i=1}^n \overline X^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n X_i^2 - 2n \overline X^2  + n \overline X^2 \right) = \cfrac{1}{n^2} \mathrm{Var}\left( \sum_\limits{i=1}^n X_i^2 - n \overline X^2 \right) =  \mathrm{Var}\left( \frac1n \sum_\limits{i=1}^n X_i^2 - \left(\frac1n \sum\limits_{i=1}^n X_i\right)^2 \right) = \mathrm{Var}(\mathsf{E}(X_i^2) - (\mathsf{E}(X_i))^2) = \mathrm{Var}(\mathrm{Var(X_i)}) = \mathrm{Var(\sigma^2) = 0}$$
Perhaps you can see why I am not satisfied with such an answer. Thanks for the help.
 A: $\frac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}$ where $\chi^2_{n-1}$ is the Chi-square
distribution with $n-1$ degrees of freedom. See here for details.
$Var(\frac{(n-1)S^2}{\sigma^2})=2(n-1)$
$Var \left( \frac{\sum(X_i-\bar{X})^2}{\sigma^2}\right)=2(n-1)$
$Var (\sum(X_i-\bar{X})^2 )=2(n-1)\sigma^4$
$Var (\frac{1}{n}\sum(X_i-\bar{X})^2 )=\frac{2(n-1)\sigma^4}{n^2}$
A: Since you are looking at the variance of the sample variance (without Bessel's correction) you are going to get moments up to the fourth order.  So your working should ultimately reduce to some linear function of the first four moments of the normal distrbution (i.e., up to its kurtosis).  If you would like to learn more about the various moments of the sample mean and sample variance, in a general setting that does not presume normality, you can find full derivations in O'Neill (2014).  The result you are trying to prove is a special case of more general results proved in that paper.
A: The problem with your deduction lies here:
$$
\mathrm{Var}\left( \frac1n \sum_\limits{i=1}^n X_i^2 - \left(\frac1n \sum\limits_{i=1}^n X_i\right)^2 \right) = \mathrm{Var}(\mathsf{E}(X_i^2) - (\mathsf{E}(X_i))^2)
$$
as $\overline{X} \ne E \left( X_i \right)$, just $E \left( \overline{X} \right) = E \left( X_i \right)$. For similar reason $\overline{X^2} \ne E \left( X_i^2 \right)$.
Hint: I would continue from there with
\begin{align}
  \frac{1}{n^2} Var \left( \sum{X_i^2} -\frac{1}{n} \left( \sum X_i \right)^2 \right) = \\
= \frac{1}{n^2} Var \left( \frac{n - 1}{n} \sum{X_i^2} - \frac{2}{n} \sum_{i < j} X_i X_j \right) = \ldots
\end{align}
