Why do you take the sqrt of 1/n for RMSE? Updated question:
Why do we use RMSE:
$$RMSE = \sqrt{\frac{1}{n}\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
Why is it not MRSE:
$$MRSE = \frac{1}{n}\sqrt{\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
I understand that other methods (e.g., MAE and MAPE) can be used as a metric for error. My question is specifically about why we use RMSE over MRSE.
Original:
Why is the equation for RMSE:
$$RMSE = \sqrt{\frac{1}{n}\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
Why is it not:
$$RMSE = \frac{1}{n}\sqrt{\Sigma_{i=1}^{n}{\Big(\hat{y}_i -y_i\Big)^2}}$$
What is the reason for taking the square root of 1/n?

The goal is to have an unbiased estimator for the error your model makes on average. Let's call that $\bar \epsilon$. Now let's see what's the relationship of the two estimators you asked about with the $\bar \epsilon$:
$\hat y_{i} - y_{i} = \epsilon_{i}$
$\frac{1}{n}\sum_{i=1}^{n}(\hat y_{i} - y_{i})^2 = \frac{1}{n}\sum_{i=1}^{n}(\epsilon_{i})^2 = \bar \epsilon^2$
thus
$ \sqrt{\frac{1}{n}\sum_{i=1}^{n}\left(\hat{y}_{i}-y_{i}\right)^{2}} \approx \bar \epsilon$
which is what we aimed for. Now let's see what the other estimator will give you:
$\frac{1}{n} \sqrt{\sum_{i=1}^{n}\left(\hat{y}_{i}-y_{i}\right)^{2}} = $
$\frac{1}{n} \sqrt{n \times \frac{1}{n}\sum_{i=1}^{n}\left(\hat{y}_{i}-y_{i}\right)^{2}} = $
$\frac{\sqrt{n}}{n} \times \bar \epsilon$
As you can see the second estimator has a bias of $\frac{\sqrt{n}}{n}$ in estimating the average error you aimed for. For example, if for a data generating process of $f(x)  = 0$ you always predict 2 then you would want the estimator to give you $\bar \epsilon = 2$ which is given by the first estimator while the second estimator (assuming n = 10) will give you ($2 \times \frac{\sqrt 10}{10}$).
