Metric Isometry is always smooth? Let $M$ be a smooth manifold. Let $d$ be any metric on $M$ which induces the topology on $M$. Let $f:(M,d) \rightarrow (M,d) $ be an isometry (in the sense of metric spaces). Is it true that $f$ must be smooth?
(if the metric $d$ is induced by some Riemannian metric $g$ then the answer is positive by the Myers–Steenrod theorem)

No. Let $(X,d')$ be a metric space and let $f \colon X \rightarrow X$ be a homeomorphism. You can define a new metric on $X$ (which you might call the pullback metric) by $d(x,y) := d'(f(x),f(y))$. The topology induced by $d$ is the same as the topology induced by $d'$ and $\varphi \colon X \rightarrow X$ is an isometry for $(X,d')$ if and only if $f^{-1} \circ \varphi \circ f$ is an isometry for $(X,d)$. This gives you a construction that modifies isometries by possibly nonsmooth maps.
To generate a counterexample, take $M = \mathbb{R}$, $d'(x,y) = |x - y|$, $f = x^3$ (a homeomorphism but not a diffeomorphism of $\mathbb{R}$ with the usual smooth structure) and $\varphi(x) = x + 1$. Then $d(x,y) = |x^3 - y^3|$ is a metric on $\mathbb{R}$ inducing the standard topology and $\tilde{\varphi}(x) = (f^{-1} \circ \varphi \circ f)(x) = \left(x^3 + 1\right)^\frac{1}{3}$ is a nonsmooth isometry of $(\mathbb{R}, d)$.
