Are all continuous bijective translations isometries?

Question

Let $(M, d)$ be a metric space. I define a translation on $M$ to be a function $f$ from $M$ to $M$ such that $d(x,f(x))=d(y,f(y))$ for all $x$ and $y$ in $M$. In a previous question, I asked if every translation was an isometry in the post Are all metric translations isometries. The answer was no, even with continuity, and even with bijectiveness. But what if we have both bijectiveness AND continuity? Is that sufficient to force the translation to be an isometry?

Answer 1

Let $M = \{-1, 0, 1, 2\}$ with the standard distance (in $\mathbb R$). Let

$$f(-1) =0\ , f(0) = -1\ , f(1) = 2\ , f(2) =1$$

Then $d(x, f(x)) = 1$ for all $x$ and $4= d(f(0), f(1)) \neq d(0, 1) = 1$