Let $M$ be a manifold of dimension $m$ without boundary and $d$ be a complete metric on it, the group of isometric automorphisms being $\mathrm{Iso}(d)$.

Suppose $d'$ is another complete metric on $M$. $\mathrm{Iso}(d)$ is said to be homotopically embedded into/isomorphic to $\mathrm{Iso}(d')$ if a suitable choice of homotopic equivalents for each $\mathrm{Iso}(d)$ element in $\mathrm{Iso}(d')$ makes up a proper embedding/isomorphism from $\mathrm{Iso}(d)$ to $\mathrm{Iso}(d')$.

$(M,d)$ is said to have trivial ends if either $M$ is compact or $\exists p \in M$ 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 $k=m$).

$(M,d)$ is said to achieve maximal metric symmetry if there exists no other complete metric $d'$ such that $\mathrm{Iso}(d)$ is homotopically embedded into $\mathrm{Iso}(d')$. Note that $M$ may have many maximal metric symmetries, e.g. the flat metric over $\Bbb{R}^2$ and the hyperbolic metric over $\Bbb{H}^2$.

Question:

1: If $M$ is smoothable, $(M,d)$ has trivial ends and achieves maximal metric symmetry, is it true that there exists a Riemannian structure $(M,g)$ achieving the same symmetry (i.e. $\mathrm{Iso}(g)$ is homotopically isomorphic to $\mathrm{Iso}(d)$)? It seems that Thurston's geometrization partly answers the 3-dimensional case, but I'm not certain.

2: If the previous conjecture is false, then how does the topology structure of $M$ influence its maximal symmetry, like all possible maximal symmetry groups being determined by $H_*(M)$? I'm especially interested in non-smoothable cases.