Maximal symmetries 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).
 A: This may be more appropriately regarded as a comment than as an answer.
I realize that this question has been around for a while, probably because it's not completely clear what some parts of the question mean.  The usage of 'homotopic subgroup' and 'isotopic subgroup' have already been pointed out as unclear, but that's not all.  For example, it's not clear what the OP means by 'converse' in the third paragraph.  Presumably, the OP means 'the corresponding statement obtained by exchanging $d$ and $d'$', but I'm not sure, as this is not the standard meaning of 'converse'.
Anyway, the following example might help to focus the OP's Question 1 a little better:  Consider $\mathrm{SU}(2)\simeq S^3$ and the $6$-parameter family $\mathcal{F}$ of left-invariant Riemannian metrics on $SU(2)$.  For the generic $g\in \mathcal{F}$, the isometry group is $3$-dimensional.  Its identity component consists of $\mathrm{SU}(2)$ itself, acting by left translations.  However, there is a $1$-parameter subfamily of $\mathcal{F}$, consisting of metrics of constant curvature, for which the isometry group is $6$-dimensional and isomorphic to $\mathrm{O}(4)$.  Thus, all the rest of the left-invariant metrics on $\mathrm{SU}(2)$ (which are all still homogeneous) do not acheive maximum symmetry in the sense that the OP seems to want, so that the answer to Question 1 would appear to be 'no'.
