The aim of this question is to investigate how topological group actions on manifolds differs 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 $\mathrm{Iso}(d')$ contains a topological subgroup isomorphic to $\mathrm{Iso}(d)$ and corresponding elements are homotopy equivalent. In other words, $\mathrm{Iso}(d)$ is isotopic to a subgroup of $\mathrm{Iso}(d')$ in $\mathrm{Aut}(M)$. If $\mathrm{Iso}(d')$ is also a homotopic subgroup of $\mathrm{Iso}(d)$, then we say that $(M,d)$ and $(M,d')$ achieve the same symmetry. If there exists no complete metric $d'$ such that $\mathrm{Iso}(d')$ has $\mathrm{Iso}(d)$ as a strict homotopic subgroup, then we say that $(M,d)$ achieves a maximal symmetry.

Q$1$: Do $\Bbb{S}^n$, $\Bbb{R}^n$ and $\Bbb{H}^n$ with their constantly curved metrics achieve maximal symmetries? According to this post, 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, homeomorphic to the interior of a compact manifold and $(M,d)$ achieves a maximal metric symmetry, is it always the case that there exists a Riemann structure $(M,g)$ achieving the same symmetry? What can we say about maximal symmetries when $M$ is not smoothable?