Divide by n, n minus 1, or n plus 1?

For $n$ i.i.d. draws from $N(\mu, \sigma^2)$, consider estimating $\sigma^2$ by $\hat\sigma^2_c = \frac{1}{c}\sum_{i=1}^n (X_i - \bar X)^2$. Three common choices of $c$: $n-1$ (unbiased), $n$ (the MLE), and a mystery third option.

Which divisor $c$ minimizes the mean squared error of $\hat\sigma^2_c$? Rank the three.