The aim of this question is to investigate how topological group actions on manifolds differs away from more rigid actions (like smooth ones).

Let $M$ be a connected, second-countable and boundless manifold of dimension $m$. Let $d$ and $d'$ be $2$ complete metrics on it, both inducing the manifold topology. The isometry group of $(M,d)$, denoted as $\mathrm{Iso}(d)$, is a topological subgroup of $\mathrm{Aut}(M)$ under the compact-open topology.

$\mathrm{Iso}(d)$ is said to be a **homotopic subgroup** of $\mathrm{Iso}(d')$ if there exists an embedding from $\mathrm{Iso}(d)$ to $\mathrm{Iso}(d')$ that sends each element to its homotopic equivalent. In other words, $\mathrm{Iso}(d)$ is isotopic to a subgroup of $\mathrm{Iso}(d')$ in $\mathrm{Aut}(M)$.

$(M,d)$ is said to be **simple at infinity** if $M$ is homeomorphic to the interior of a compact manifold. It is said to achieve **maximal symmetry** if there exists no other complete metric $d'$ such that $\mathrm{Iso}(d')$ has $\mathrm{Iso}(d)$ as a **strict** homotopic subgroup.

**Q$1$**: Do $\Bbb{S}^n$, $\Bbb{R}^n$ and $\Bbb{H}^n$ with their constantly curved metrics achieve maximal symmetries? According to the second answer of [this question](https://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces), flat metric of $\Bbb{R}^n$ achieves the unique maximal symmetry among normable metrics. However it seems much more complicated when it comes to non-normable cases (e.g. $\Bbb{R}^n \cong \Bbb{H}^n$ but they have incommensurable symmetries).

**Q$2$**: If $M$ is smoothable, $(M,d)$ is simple at infinity and achieves maximal metric symmetry, is it true that there exists a Riemann structure $(M,g)$ achieving the same symmetry (i.e. $\mathrm{Iso}(g)$ is homotopically isomorphic to $\mathrm{Iso}(d)$)? What can we say about maximal symmetries when $M$ is not smoothable?