Metric on $\Bbb{R}$ Does there exist a complete metric $d$ on $\Bbb{R}$ which induces the manifold topology that has an orientation-preserving isometry group strictly larger than / incomparable to $(\Bbb{R},+)$?
I want to know how well the symmetry of a manifold (with as few as possible structures) can be approached by more rigid structures, like smooth (therefore Riemann) structures. $\Bbb{R}$ is chosen because of its simplicity. A further question (and still unanswered) is asked here.
 A: First, some background. This type of questions belongs to the area of mathematics known an "Topological transformation groups." This area was quite popular until 1960s and its stems from Hilbert's 5th problem about characterization of Lie groups. The best book on the subject still is the one originally published in 1955:
Montgomery, Deane; Zippin, Leo, Topological transformation groups, Mineola, NY: Dover Publications (ISBN 978-0-486-82449-9). xi, 289 p. (2018). ZBL1418.57024.
The hardest open problem in the theory is known as Hilbert-Smith Conjecture (HSC):
Suppose that $G$ is a (Hausdorff) locally compact topological group acting continuously and effectively on a topological manifold $M$. Then $G$ isomorphic (as a topological group) to a Lie group.
In view of the structural results on locally compact groups,  this reduces to the case when $G$ is a compact group. A continuous action of $G$ on $M$ then has an invariant metric which metrizes the standard topology of $M$. Thus, the assumption of existence of an invariant metric does not make any difference in this setting.
The best (to my knowledge) background discussion of HSC can be found in Terry Tao's blog, here. The problem is known to reduce to the one when $G$ is compact and isomorphic (as a topological group) to the group of $p$-adic integers (for various values of $p$).
HSC is known to hold if $M$ is 1-dimensional (see below) and 2-dimensional (see the above book). Three-dimensional case was settled relatively recently (2013), by John Pardon.
The conjecture is wide-open in dimensions $\ge 4$.
Let me now address your question (I will not be using the known result about HSC but give a direct proof). Suppose that $G$ is the isometry group of a metric $d$ on the real line $L$, as in your question, $G_+<G$ is the subgroup of orientation-preserving isometries. I will equip $G$ with the topology of uniform convergence on compacts (equivalently, compact-open topology). Then an application of a version of the Arzela-Ascoli theorem implies that $G$ is locally-compact. The stabilizer $G_x< G$ of any point $x\in L$ is a compact subgroup of $G$.
The following lemma is elementary:
Lemma 1. Suppose that $H< G$ is a subgroup and $x$ is a point in $L$. Then the orbit  $Hx$ is either discrete (i.e. is a discrete when equipped with the subspace topology) or contains $x$ in its closure.
Lemma 2. The subgroup $G_+< G$ acts freely on $L$.
Proof. Suppose that a point $x\in L$ has nontrivial stabilizer in $G_+$, let $g$ be a nontrivial element of this stabilizer. Then the subgroup $H$  generated by $g$ is relatively compact. Let $F\subset L$ denote the fixed-point set of $g$; then $F\ne L$, since $g\ne 1_G$. Pick a complementary component $J$ of $F$ in $L$. Then, after replacing $g$ with $g^{-1}$ if necessary, for each $y\in J$ we have
$$
g(y)> y. 
$$
But then the sequence of iterates $(g^n(y))_{n\in {\mathbb N}}$ cannot converge to $y$, contradicting Lemma 1. qed
Suppose now that $G$ contains a connected nontrivial subgroup $H$, e.g. one isomorphic to ${\mathbb R}$ as in your question.
Lemma 3. Under the above assumption, $G_+=H$.
Proof. Take any $x\in L$ and consider the orbit map
$$
o_x: G_+\to L, o_x(g)=g(x). 
$$
Lemma 2 implies that this map is a continuous injective map. The image $o_x(H)$ is connected (since $H$ is), hence, is an open interval $J\subset L$. It suffices to prove that $J=L$. Suppose not. Then, since $H$ is connected, it cannot switch the boundary points of $J$ (if there are two), hence, has to fix both boundary points of $J$, contradicting Lemma 1.
Thus, $H$ acts transitively on $L$. Thus, in view of Lemma 1, the action of $G_+$ on $L$ is simply-transitive, which implies that $G_+=H$. qed
Corollary 1. If $G_+$ contains a subgroup $H$ isomorphic to ${\mathbb R}$ as a topological group, then $G_+=H$.
One actually can do better:
Lemma 4. Suppose that $G$ contains an uncountable closed subgroup $H$. Then $H$ acts transitively on $L$.
Proof. Since $G_+$ acts on $L$ isometrically, the action is also proper, implying that each orbit map $o_x: H\to L$ has closed image $Hx\subset L$. Suppose that the image is not equal to $L$. Since we assumed that $H$ is uncountable, so is its orbit $Hx\subset L$. Then the orbit $Hx$ contains at least one point which is a boundary point of one of the arcs which are components of $L\setminus Hx$ and also one point which is not such a boundary point. This contradicts transitivity of the action of $H$ on the orbit $Hx$. qed
Corollary 2. The subgroup $H$ as in Lemma 4 is homeomorphic to ${\mathbb R}$.
Proof. Since the action of $H$ on $L$ is proper, the orbit map $o_x: H\to L$ is a homeomorphism. qed
Now, I will simply quote answers to this question to conclude that a subgroup $H$ as in Corollary 2 is isomorphic to ${\mathbb R}$ as a topological group. This proves:
Theorem. Let $(L,d)$ is a metric space homeomorphic to ${\mathbb R}$, $G$ is the isometry group of $(L,d)$ and $G_+< G$ is the orientation-preserving subgroup. Then either $G_+$ is at (most countable) or is isomorphic to ${\mathbb R}$ as a topological group.
In fact, in the countable case, $G_+$ is either trivial or infinite cyclic. I will leave you a proof as an exercise.
