It is well known that $C^1$ (actually even just differentiable) isometries of Riemannian manifolds are actually $C^\infty$. The proof is based on the metric structure generated by the Riemannian metric, see e.g. A differentiable isometry is smooth?.
On the other hand, the following paper https://arxiv.org/pdf/1311.0199.pdf shows that a $C^2$ isometry of a pseudo-Finsler structure is in fact $C^\infty$ using that the pseudo-Finsler defines a connection on the double tangent bundle.
For a $C^\infty$ pseudo-Riemmanian metric $g$ on a closed manifold $M$, is there any reason to expect that a $C^1$ isometry, i.e., a $C^1$ diffeomorphism $f:M \to M$ such that $g_{f(p)}(Df(v), Df(w)) = g_p(v, w)$ for all $p \in M$ and $v, w \in T_p M$, should have higher regularity, for example $C^\infty$?
