If $K=1$, you can even have the stronger condition $d(f(x),f(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 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.