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 the converse proposition also holds, then we say that $(M,d)$ and $(M,d')$ **achieve the same symmetry**. If the converse proposition never holds in the strict sense, then we say that $(M,d)$ **achieves a maximal symmetry**. 

**Q$1$**: Do [homogeneous spaces](https://mathworld.wolfram.com/HomogeneousSpace.html), equipped with an arbitrary invariant smooth metric, always achieve maximal symmetries? I'm particularly interested in the constant-curvature case. [A previous post](https://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces) 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).