Let $(X, d)$ be a locally compact metric space. Define $\mathrm{Iso}(X)$, the isometry group of $X$, as the set of all surjective isometries of $X$. Is it true that $\mathrm{Iso}(X)$ is locally compact when endowed with the compact-open topology (or, equivalently, the topology of uniform convergence on compact subset of $X$)?
I am not particularly sure how to see if this is true or not. I know that if $X$ is a proper metric space (i.e. closed balls are compact) then $\mathrm{Iso}(X)$ is indeed locally compact. This is because, for fixed $x \in X$ and $r > 0$, we can consider the neighbourhood $$ N_r := \{f \in \mathrm{Iso}(X) \ \mid \ f(\overline{B}(x, r)) \subset \overline{B}(x, 2r) \} $$ of the identity map on $X$, and one can quickly show that this set is relatively compact using the Arzela-Ascoli theorem. The hard part of the proof is to show that for every $y \in X$, the set $$ \{f(y) \ \mid \ f \in N_r \} $$ is relatively compact in $X$. However, for fixed $z \in \overline{B}(x, r)$, we have that $$d(f(y), x) \leq d(f(x), f(z)) + d(f(z), x) \leq d(x, z) + 2r, $$ for every $f \in N_r$, and so relative compactness follows immediately. The same argument works for an arbitrary isometry (not necessarily the identity map).
However, if $X$ is only locally compact, then we know that each point in $X$ has arbitrarily small closed balls that are compact, but arbitrarily big closed balls may not necessarily be compact, which is why the above proof does not directly work in this case.
For a simple counterexample, let $X$ be an infinite set with the discrete metric $d(x,y)=1$ if $x\neq y$. Then $\mathrm{Iso}(X)$ is just the set of all bijections $X\to X$ and the compact-open topology is the product topology. A basic open set is then of the form $U=\{f\in\mathrm{Iso}(X):f(x_1)=y_1,\dots,f(x_n)=y_n\}$ for some $x_1,\dots,x_n,y_1,\dots,y_n\in X$ (with the $x_i$ all distinct and the $y_i$ all distinct). If you pick $x\in X\setminus\{x_1,\dots,x_n\}$ then the evaluation map at $x$ maps $U$ onto $X\setminus\{y_1,\dots,y_n\}$, since an element of $U$ can map $x$ anywhere other than $y_1,\dots,y_n$. Since $X\setminus\{y_1,\dots,y_n\}$ is not precompact in $X$ and the evaluation map is continuous, this means $U$ is not precompact. So no compact subset of $X$ can contain any basic open set, so no compact subset of $X$ has nonempty interior.