I know that if the sample variance of a random sample from a normal distribution $(\mu,\sigma^2)$ is
$$S_1^2 = \frac{1}{n-1}\sum{(X_i-\bar{X})}^2$$
then,
$$U =\frac{(n-1)S_1^2}{\sigma^2}$$ has a $\chi^2$ distribution with $n-1$ degrees of freedom.
Does this mean that if my sample variance of a random sample from a normal distribution $(\mu,\sigma^2)$ is
$$S_2^2=\frac{1}{n}\sum{(X_i-\bar{X})}^2$$
then,
$V = \dfrac{nS_2^2}{\sigma^2}$, has a $\chi^2$ distribution with $n$ degrees of freedom?
No. Not quite. In general, the sample variance $$ S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2, $$ given $X_i\overset{\mathrm{iid}}{\sim}\mathcal N(\mu,\sigma^2)$, is distributed according to $S^2\sim\operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2\sigma^2}{n-1})$ (shape-scale parameterization). Through a slight abuse of notation we write $$ S^2\sim \operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2\sigma^2}{n-1}) $$ $$ \tfrac{n}{\sigma^2}S^2\sim \tfrac{n}{\sigma^2}\operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2\sigma^2}{n-1}) $$ $$ \tag{1} \tfrac{n}{\sigma^2}S^2\sim\operatorname{Gamma}(\tfrac{n-1}{2},\tfrac{2n}{n-1}). $$ The right hand side of $(1)$ cannot be written as a simple $\chi^2$-distribution as it is equivalent to $$ \tfrac{n}{\sigma^2}S^2\sim\tfrac{n}{n-1}\underbrace{\operatorname{Gamma}(\tfrac{n-1}{2},2)}_{\chi^2(n-1)}, $$ which is to say $\tfrac{n}{\sigma^2}S^2$ has the same distribution as a $\chi^2(n-1)$ random variable multiplied by $n/(n-1)$.