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

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

We say that $\mathrm{Iso}(d')$ is a homotopic subgroup of $\mathrm{Iso}(d)$, if $\mathrm{Iso}(d)$ contains a topological subgroup isomorphic to $\mathrm{Iso}(d')$ and the corresponding elements under that isomorphism are homotopic maps from $M$ to itself. In other words, $\mathrm{Iso}(d')$ is isotopic to a subgroup of $\mathrm{Iso}(d)$ in the topological space $\mathrm{Aut}(M)$.

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 the $\mathrm{Iso}(d)$ is never a strict homotopic subgroup of $\mathrm{Iso}(d')$ for any $d'$, then we say that $(M,d)$ achieves a maximal symmetry.

Q$1$: Do homogeneous spaces, equipped with an arbitrary invariant smooth metric, always achieve maximal symmetries? I'm particularly interested in the constant-curvature case. A previous post says that the homogeneous flat metric on $\Bbb{R}^n$ achieves the unique maximal symmetry among normable metrics. However it seems much more complicated when considering non-normable cases (e.g. $\Bbb{R}^n$ is diffeomorphic to $\Bbb{H}^n$ but their symmetries are incommensurable).

Q$2$: Can maximal symmetries on smooth manifolds always be realized by smooth structures (i.e. if $(M,d)$ achieves a maximal symmetry, then there exists a Riemann structure $(M,g)$ achieving the same symmetry)? I'm particularly interested in the simple-at-infinity cases (i.e. $M$ is homeomorphic to the interior of a compact manifold).