 Solution

Take $M=\mathbb{R}$ and the constant contraction mapping $f(x)=0$. For $p=(0,0)$, $q=(0,1)$,No Lipschitz constant less than one can work, so $F$ need not be a contraction. More generally, differing only in the second input coordinate already prevents a strict contraction whenever $M$ has two points.

If $f$ has contraction constant $0\le c<1$, two iterations giveand consequentlyThus $F^2$ is a contraction mapping. The swap lift of a contraction shows why an iterate can contract even when one step does not.