Let $X_1, X_2, \dots, X_n$ be iid random variables. Consider the sample variance $$S_n^2:=\frac{1}{n-1} \sum_{i=1}^{n}(X_i - \overline{X})^2$$ where $\overline{X}:=\frac{1}{n}\sum_{i=1}^n X_i$ is the sample mean. If $S_n^2$ is consistent estimator of $\sigma^2$; i.e., $S_n^2 \overset{P}{\to} \sigma^2$ as $n \to \infty$. I wonder what would be the limit of the scaled sample variance; i.e., $\frac{S_n^2}{2} \overset{P}{\to} ?$.
On the one hand, by a continuous mapping theorem, we know that $g(S_n^2) \to g(\sigma^2)$ for some continuous function $g$. Thus, it follows that $\frac{S_n^2}{2} \overset{P}{\to} \sigma^2/2.$ However, the point that I do not understand is that why not $\frac{S_n^2}{2} \overset{P}{\to} \frac{\sigma^2}{2^2}$? The idea behind this thinking is that for $a \in \mathbb{R}$ and a random variable $X$, ${\rm var}(a X) = a^2 {\rm var}(X)$---not $a {\rm var}(X)$.
Did I misunderstand the theorem or I made the mistake somewhere.