Big-O notation compares how algorithms scale, so an $O(n\log n)$ algorithm always runs faster than an $O(n^{2})$ one on the same input.

False

False. Big-O describes growth as n tends to infinity and discards constant factors, so a heavily-optimised $O(n^{2})$ routine can beat a constant-heavy $O(n\log n)$ one at every n a real program ever sees. The crossover point is a measurement, not something the notation tells you.