The unbiased sample variance is not the MSE winner

Let $X_1,\ldots,X_n$ be independent $N(\mu,\sigma^2)$ with both parameters unknown and $n\ge2$. Write $Q=\sum_{i=1}^n(X_i-\bar X)^2$ and $S^2=Q/(n-1)$. Under squared-error loss for the target $\sigma^2$, compute the exact risk of $S^2$ and of the MLE $\hat\sigma^2=Q/n$. Then, for a constant $a\ge0$, compute the exact risk of $aS^2$ and find the unique $a^*$ minimizing that risk. Identify the corresponding estimator as a multiple of $Q$ and compute its risk. Prove that $$ R(\sigma^2,S^2) >R(\sigma^2,\hat\sigma^2) >R(\sigma^2,a^*S^2) $$ for every $\sigma^2>0$. This is a comparison of estimators of a scale parameter. Do not compare linear estimators $aX$ of a normal mean, and do not invoke a decision-theoretic shrinkage argument for $\mu$.

Unbiasedness of $S^2$ makes its risk equal to its variance.