Maximal symmetries of complete metrics on manifolds

Question

Let $M$ be a connected and second-countable manifold, and $d,d'$ be $2$ complete compatible metrics on it. The isometry group of $(M,d)$, denoted as $\mathrm{Iso}(d)$, is equipped with the compact-open topology.

We say that $\mathrm{Iso}(d')$ is a homotopic subgroup of $\mathrm{Iso}(d)$, if there is a topological group embedding $\sigma : \mathrm{Iso}(d') \rightarrow \mathrm{Iso}(d)$ such that $\forall g \in \mathrm{Iso}(d')$, $g$ is homotopic to $\sigma(g)$ as self maps of $M$.

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 $\mathrm{Iso}(d)$ 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? A previous post says that the homogeneous flat metric on $\Bbb{R}^n$ achieves the unique maximal symmetry among normable metrics. However maximal symmetry may not be unique, e.g. $\Bbb{R}^n$ is diffeomorphic to $\Bbb{H}^n$ but their symmetries are incomparable. Edit: This is answered negatively by Robert Bryant.

Q$2$: Can maximal symmetries on compact smoothable manifolds always be realized smoothly (i.e. if $(M,d)$ achieves a maximal symmetry, then there exists a Riemann structure $(M,g)$ achieving the same symmetry)?

Answer 1

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'.