The MLE beats the UMVUE in MSE for normal variance estimation

Suppose $\{X_1,X_2,\dots,X_n\}$ is a random sample of size $n$ from the $N(\mu,\sigma^2)$ distribution, $\mu\in\mathbb{R}, \sigma^2>0$. Suppose $\hat\sigma_n^2$ and $T_n$ denote the maximum likelihood estimator and the UMVUE of $\sigma^2$ respectively. Which of the following statements is correct?