$(M,d)$ is a compact metric space and $f:M \to M$ is bijective such that $d(f(x),f(y)) \le d(x,y) , \forall x,y \in M$ , then is $f$ an isometry?

Question

$(M,d)$ is a compact metric space and $f:M \to M$ is an bijective function such that $d(f(x),f(y)) \le d(x,y) , \forall x,y \in M$ , then is $f$ an isometry i.e. $d(f(x),f(y)) = d(x,y) , \forall x,y \in M$ ?

Answer 1

I guess that $x\mapsto \frac{1}{2} x$ from $[0,1]$ into intself is injective, but is not an isometry. The answer is no.