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.