What happens to the running time of an $O(n \ln n)$ algorithm if you double $n$?
I  suggest following property: if for functions $g,f \gt 0$ we have $f \sim kg$, for some constant $k\gt 0$  i.e. $\lim\limits_{n \to \infty}\frac{f(n)}{g(n)}=k$ and both, $g$ and $f$, have not zero as limit point, then $O(f)=O(g)$.
Proof:
Let's consider $\phi \in  O(f)$, then we have $\phi \leqslant C f \leqslant C (k+\varepsilon)g$ for appropriate $C, \varepsilon$ and $n \gt N.$ So $O(f) \subset O(g)$. Using, that we have also $\lim\limits_{n \to \infty}\frac{g(n)}{f(n)}=\frac{1}{k} \gt 0$, we obtain reverse including.
Based on brought property we can say, that running time $O(n \ln n)$ does not change if we double $n$:
$$O(n \ln n) = O(2n \ln 2n)$$