Just a quick counterexample to your last question:

Let $C \subset \mathbb{R}$ be a fat Cantor set that is symmetric w.r.t. $0$ and define $f(x) = \begin{cases}x,& \text{if }x \in C\\\ -x,& \text{if }x \notin C \end{cases}.$

(More trivial counterexample would be $f(x) = \begin{cases}x,& \text{if }x \notin \mathbb{Q}\\\ 0,& \text{if }x \in \mathbb{Q} \end{cases}$, but changing things in a set of measure zero is not that satisfying.)

Edit: Perhaps even more satisfying example would be something like $f(\sum_{i=0}^\infty x_i2^{-i}) = \sum_{k=0}^\infty (1-x_{2k})2^{-2k}+\sum_{k=0}^\infty (1-x_{2k+1})2^{-2k-1}$ with $x_i \in \{0,1\}$, i.e. switch every other digit in the binary representation (does not matter what you do at the endpoints of intervals).

Just a quick counterexample to your last question:

Let $C \subset \mathbb{R}$ be a fat Cantor set that is symmetric w.r.t. $0$ and define $f(x) = \begin{cases}x,& \text{if }x \in C\\\ -x,& \text{if }x \notin C \end{cases}.$

(More trivial counterexample would be $f(x) = \begin{cases}x,& \text{if }x \notin \mathbb{Q}\\\ 0,& \text{if }x \in \mathbb{Q} \end{cases}$, but changing things in a set of measure zero is not that satisfying.)

Edit: Perhaps even more satisfying example would be something like $f(\sum_{i=0}^\infty x_i2^{-i}) = \sum_{k=0}^\infty (1-x_{2k})2^{-2k}+\sum_{k=0}^\infty x_{2k+1}2^{-2k-1}$ with $x_i \in \{0,1\}$, i.e. switch every other digit in the binary representation (does not matter what you do at the endpoints of intervals).

Just a quick counterexample to your last question:

Let $C \subset \mathbb{R}$ be a fat Cantor set that is symmetric w.r.t. $0$ and define $f(x) = \begin{cases}x,& \text{if }x \in C\\\ -x,& \text{if }x \notin C \end{cases}.$

(More trivial counterexample would be $f(x) = \begin{cases}x,& \text{if }x \notin \mathbb{Q}\\\ 0,& \text{if }x \in \mathbb{Q} \end{cases}$, but changing things in a set of measure zero is not that satisfying.)

Just a quick counterexample to your last question:

Let $C \subset \mathbb{R}$ be a fat Cantor set that is symmetric w.r.t. $0$ and define $f(x) = \begin{cases}x,& \text{if }x \in C\\\ -x,& \text{if }x \notin C \end{cases}.$

(More trivial counterexample would be $f(x) = \begin{cases}x,& \text{if }x \notin \mathbb{Q}\\\ 0,& \text{if }x \in \mathbb{Q} \end{cases}$, but changing things in a set of measure zero is not that satisfying.)

Edit: Perhaps even more satisfying example would be something like $f(\sum_{i=0}^\infty x_i2^{-i}) = \sum_{k=0}^\infty (1-x_{2k})2^{-2k}+\sum_{k=0}^\infty (1-x_{2k+1})2^{-2k-1}$ with $x_i \in \{0,1\}$, i.e. switch every other digit in the binary representation (does not matter what you do at the endpoints of intervals).