Normalizer of translation subgroup in homeomorphism group Edit: Could someone check my solution below, so I can accept the answer and complete this post. Thanks!
I want to find all homeomorphism $g :\mathbb{R} \rightarrow \mathbb{R}$ for which $$g\circ H \circ g^{-1} = H$$
where $H$ is the translation subgroup.  Given $\tau_t (x) = x-t$, we want
$$g\circ\tau_t\circ g^{-1} = \tau_h$$
for some $\tau_h \in H$.
Writing it out explicitly 
$$g(g^{-1}(x) - t) = x-h$$
this implies
$$g^{-1}(x) - t = g^{-1}(x-h)$$
but I couldn't find a way to find all possible solution of this functional equation. 
I also know the subgroup of all affine functions $ax+b$ will certainly work. For fixed $t$,$h$, there are solutions to the above functional equation which is not in $ax+b$ form. But I think this will not be a problem since those solutions will not work if we vary $t$.
 A: The key is that all continuous automorphisms of the topological group $({\mathbb R}, +)$ are linear. (This is false for arbitrary automorphisms, of course.) Proving this lemma is a nice exercise. (Hint: First consider automorphisms of the group of rational numbers.) Use this lemma for the continuous automorphism of ${\mathbb R}$ given by conjugation via $g\in   N({\mathbb R})$. Then for each $g\in N({\mathbb R})$, there exists $a\ne 0$ such that for every $x, t\in {\mathbb R}$ we have 
$$
g^{-1}(x+t)- g^{-1}(x)= at.
$$
From this it follows that $h=g^{-1}$ has derivative equal to $a$ at every $x\in {\mathbb R}$, from which it follows that $g(x)= a^{-1}x + b$, where $b$ is constant. 
A: Given $T:= \{\tau_r: r\in \mathbb{R}\}$, we will first show that the each normalizer defines a continuous automorphism on $\mathbb{R}$.
Give $g\in N(T)$, define the map $G:\mathbb{R} \rightarrow \mathbb{R}$ by $G(r) = h$ where $h$ is the sub index $g\tau_r g^{-1} = \tau_h$.


*

*$G$ is well defined, since each $r$ gives a $\tau_r$

*$G$ is onto, given $l\in \mathbb{R}$, have $g^{-1}\tau_l g = \tau_r$ thus $G(r) = l$

*$G$ is one-to-one since if $r_1\neq r_2 \Longrightarrow \tau_{r_1} \neq \tau_{r_2} \Longrightarrow g\tau_{r_1} g^{-1} \neq g\tau_{r_2}g^{-1}$ 

*$G$ is a homomorphism since $\tau_{G(r+l)} = g\tau_{r+l} g^{-1} = g\tau_r \tau_lg^{-1}=g\tau_r g^{-1}g \tau_lg^{-1} = \tau_{G(r)}\tau_{G(l)} = \tau_{G(r)+ G(l)}$

*$G$ is continuous, since given $r_n \rightarrow r$, we have $\tau_{G(r_n)} = g\tau_{r_n} g^{-1} = g(g^{-1}(x)+r_n) \rightarrow g(g^{-1}(x)+r) = g\tau_{r} g^{-1} = \tau_{G(r)}$ because $g$ is continuous.


Now to show all continuous automorphisms on $\mathbb{R}$ is linear function, note that for each rational number, we have 
$$G(p/q) = p G(1/q) = (p/q) q G(1/q) = p/qG(1)$$
by continuity, it has to be a linear function on $\mathbb{R}$. 
Now if $G(r) = ar$, we look back to what $g$ could be. Given 
$$g\tau_{r}g^{-1} = \tau_{ar}$$
this implies 
$$g^{-1}(x) + r = g^{-1}(x+ar) $$
or 
$$g^{-1}(x+l) - g^{-1}(x) = l/a$$
divide by $l$ and send $l$ to zero we have the derivative of $g^{-1}$ is $1/a$ at all $x$, and therefore $g = ax+b$ for some constant $b$.  
