I'm trying to calculate the variance of the sample variance of a normal distribution. Let $Y_1,Y_2,...,Y_n$ be a sample of size $n$ from a normal distribution with mean $\mu$ and variance $\sigma^2$. Define the unbiased sample variance as $S^2=\frac1{n-1}\sum_{i=1}^{n}(\bar Y-Y_i)^2$ where $\bar Y$ is the sample mean. As we know, $\frac{(n-1)S^2}{\sigma^2}$ is $ \chi^2$ distributed with $n-1$ degrees of freedom, so $$Var(S^2)=\frac{\sigma^4}{(n-1)^2}Var(\frac{(n-1)S^2}{\sigma^2})=\frac{2\sigma^4}{n-1}$$However I am trying to calculate it in a different way.
Notice that $\bar Y$ is normal with mean $\mu$ and variance $\frac {\sigma^2}n$. Each $Y_i$ is normal with mean $\mu$ and variance $\sigma^2$. It follows that $\bar Y-Y_i$ is normal with mean $0$ and variance $\frac {\sigma^2}n+\sigma^2=\frac{(n+1)\sigma^2}{n}$. Therefore $\frac{\bar Y-Y_i}{\sqrt{\frac{(n+1)\sigma^2}{n}}}$ is a standard normal variable. Using this and the fact that $Z^2=\chi^2$ I get that: $$Var(S^2)=\frac 1{(n-1)^2}Var(\sum_{i=1}^{n}(\bar Y-Y_i)^2)=\frac 1{(n-1)^2}Var(\sum_{i=1}^{n}(\frac{\bar Y-Y_i}{\sqrt{\frac{(n+1)\sigma^2}{n}}})^2\times\frac{(n+1)\sigma^2}{n})$$ $$=\frac {(n+1)^2\sigma^4}{n^2(n-1)^2}Var(\sum_{i=1}^{n}Z_i^2)=\frac {2n(n+1)^2\sigma^4}{n^2(n-1)^2}=\frac {2(n+1)^2\sigma^4}{n(n-1)^2}$$ since $\sum_{i=1}^{n}Z_i^2$ is $\chi^2$ distributed with $n$ degrees of freedom. My derivation of the variance seems correct to me but the answer is clearly not, where did I go wrong?