The aim of this question is to find some intrinsic topological symmetry of manifolds. For example, the $3$-torus $\Bbb{T}^3$ is intuitively somewhat more symmetric than $\Bbb{T}^3 \# \Bbb{T}^3$, and this is partly supported by the fact that the absolute mapping degree of a continuous map onto itself can be arbitrary large for $\Bbb{T}^3$ but not exceeding $1$ for $\Bbb{T}^3 \# \Bbb{T}^3$. However this topic is subtle, cause there does exist $3$-manifold $K$ such that a continuous map from $K$ to $\Bbb{T}^3 \# \Bbb{T}^3$ can have arbitrary large absolute mapping degree. So I turned for a slightly more rigid structure, namely the metric structure, to study the symmetry.

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 $(M,d)$ is denoted as $\mathrm{Iso}(d)$, which is a subgroup of $\mathrm{Aut}(M)$.

$\mathrm{Iso}(d)$ is said to be homotopically embedded into/isomorphic to $\mathrm{Iso}(d')$ if there exists a proper embedding/isomorphism from $\mathrm{Iso}(d)$ to $\mathrm{Iso}(d')$ that sends each element to its homotopic equivalent.

$(M,d)$ is said to be simple at infinity if $\exists p \in M$ and $r \in \Bbb{R}^+$, each connected component of $\{q \in M|d(q,p)>r\}$ is homeomorphic to some $N \times \Bbb{R}^k$, where $N$ is a closed manifold of dimension $m-k$ (a single point when $k=m$, and empty when $M$ is compact). In this situation $M$ is always homeomorphic to the interior of a compact manifold.

$(M,d)$ is said to achieve maximal metric symmetry if there exists no other complete metric $d'$ such that $\mathrm{Iso}(d)$ can be homotopically embedded into $\mathrm{Iso}(d')$.

Question:

1: Does a maximal metric symmetry actually exist for $M$? Specifically, Does $\Bbb{R}^n$, $\Bbb{S}^n$ and $\Bbb{H}^n$ with their constantly curved metrics achieve maximal metric symmetries?

2: If $M$ is smoothable, $(M,d)$ is simple at infinity and achieves maximal metric symmetry, is it true that there exists a Riemann structure $(M,g)$ achieving the same symmetry (i.e. $\mathrm{Iso}(g)$ is homotopically isomorphic to $\mathrm{Iso}(d)$)? What can we say about maximal symmetries when $M$ is not smoothable?