By Banach fixed point theorem, if a metric on a metric space $X$ is such that $d(f(x),f(y))\leq K d(x,y)$ for $K\in (0,1)$ then $f$ has one unique fixed point.
Is there an example where $d(f(x),f(y))\leq K d(x,y)$ does not have a fixed point if $K=1$?
What if $X$ is a compact space?
On
If $K=1$, you can even have the stronger condition $d(f(x),f(y))<Kd(x,y)$ without having a fixed point (provided the space is not compact). For example, let the space be the real line with the usual metric, and let $f(x)=\sqrt{x^2+1}$.
Later note: This next paragraph has problems, but I suspect it can be fixed by adding something.....
If the space is compact, then $d(f(x),f(y))/d(x,y)$ must attain its maximum value $C$, which must be less than $1$. Then we have $d(f(x),f(y))\leq Cd(x,y)$ with $0<C<1$ so you'd have the usual conclusion about contraction mappings having a unique fixed point, which his attractive, if it's a complete metric space.
For $X=\mathbb{R}$, $d$ the usual metric, and $f(x)=x+1$, we have $d(f(x),f(y)) = |(x+1)-(y+1)| = |x-y| = d(x,y)$, so it satisfies the desired inequality with $K=1$, but $f$ clearly has no fixed point.
For a compact example, take $X=S^1$ with the distance inherited by embedding it in $\mathbb{R}^2$ as the unit circle, and $f$ a rotation that is not by an angle that is a multiple of $2\pi$. Again, $d(f(x),f(y))=d(x,y)$ for all $x$ and $y$ , but $f$ has no fixed points.