Q1) May I know why for some Markov chains like a Markov chain on a line segment of length $n$, say started at $0$, the chain mixes by the time it reaches it's diameter i.e. state $n$ (which takes $O(n^2)$ time by Central limit theorem)? I am looking for an intuition as to why reaching the diametrically opposite point guarantees mixing.
Q2) For a Markov chain on a hypercube, say started at $[0,\cdots, 0]$, if it mixes by the times it reaches it's diametrically opposite point $[1,\cdots,1]$ then, because I know that the mixing time of a Markov chain on a hypercube is $O(n\log n)$, can I think of the Markov chain on a hypercube as less diffusive than one on a line segment because the chain on the hypercube is taking $O(n\log n)$ steps to reach a distance $n$ as opposed to $n^2$ steps on a line segment? This, of course, assumes that the chain mixes by the times it reaches it's diametrically opposite point.
Any ideas?