Given a map $F:X \to X$ on a complete metric space $(X,d)$, and let $K<1$ such that:
$$ d(F(x), F(y)) \le K d(x,y), \quad \forall x,y \in X $$
then the contraction mapping theorem tells us that $F$ has a unique fixed point, and we can iteratively solve for this fixed point.
My question is, if we take $K=1$, then it is no longer a contraction, I've seen this being called a 'non-strict' contraction. I'm wondering if there are any results regarding this case and fixed points? Do they exist but aren't unique, or do they not exist at all?
In general, they do not need to admit fixed points. Take for example any translation (different from the identity) in $\mathbb{R}^n$. It has Lipschitz constant equal to 1, but has no fixed point.
On the other hand, the identity also has Lipschitz constant equal 1, and there are quite a lot of fixed points.
Let $X$ any set and endow it with the discrete metric, then any injective map $f: X \rightarrow X$ has Lipschitz constant equal 1. Thus, you can have any number of fixed points you like (just fix the points of your choice and make sure that you have an bijection without fixed points on the complement. You can actually do that, see for example here Existence of a bijective function with no fixed points).