The chi-squared distribution with $n$ degrees of freedom is defined as the distribution of a sum of squares of $n$ independent standard normal random variables $Z_i$. So the most basic defining result for chi-squared random variables is that for $X_1, ..., X_n \text{ ~ IID N}(\mu, \sigma^2)$ we have:

$$\sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2 = \sum_{i=1}^n Z_i^2 \text{ ~ } \chi_n^2.$$

Multiplying both sides by $\sigma^2/n$ gives the result:

$$\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \text{ ~ } \frac{\chi_n^2 \cdot \sigma^2}{n}.$$

Yes, it's true: The chi-squared distribution with $n$ degrees of freedom is defined as the distribution of a sum of squares of $n$ independent standard normal random variables $Z_i$. So the most basic defining result for chi-squared random variables is that for $X_1, ..., X_n \text{ ~ IID N}(\mu, \sigma^2)$ we have:

$$\sum_{i=1}^n \left(\frac{X_i - \mu}{\sigma}\right)^2 = \sum_{i=1}^n Z_i^2 \text{ ~ } \chi_n^2.$$

Multiplying both sides by $\sigma^2/n$ gives the result:

$$\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \text{ ~ } \frac{\chi_n^2 \cdot \sigma^2}{n}.$$