Recall that $X\subseteq\mathbb{R}$ is order-isomorphic to $\mathbb{R}$ whenever there is an bijective map $j:X\to\mathbb{R}$ such that $j(x)<j(y)$ if and only if $x<y$.
Conjecture 1. If $X\subseteq\mathbb{R}$ and $X$ is order-isomorphic to $\mathbb{R}$, then $X$ is the union of disjoint intervals of $\mathbb{R}$ of positive measure.
This seems to strong to be true. On the other hand, it seems like if it were false then it should be easy to produce a counter-example, which I have as yet been unable to do.
I might be able to get by with the following, weaker conjecture.
Conjecture 2. If $X\subseteq\mathbb{R}$ and $X$ is order-isomorphic to $\mathbb{R}$, then $X$ is the union of disjoint intervals of $\mathbb{R}$ of positive measure, as well as a set of measure zero.
Let $K\subset\mathbb{R}$ be a closed, nowhere dense set of positive measure, and let $U=\mathbb{R}{\setminus}K$.
Write $U=\bigcup_{n=1}^\infty U_n$ as a union of nonempty, pairwise disjoint open intervals.
Let $V=\bigcup_{n=1}^\infty V_n$, where $V_n$ is the open subinterval of $U_n$ with the same center, but half the length.
Explicitly, if $U_n=(a_n,b_n)$, let $V_n=(c_n,d_n)$, where \begin{align*} c_n&=\frac{3a_n+b_n}{4}\\[4pt] d_n&=\frac{a_n+3b_n}{4}\\[4pt] \end{align*} Now let $X=K\cup V$.
Clearly $X$ cannot be expressed as the union of a measure zero set and a union of intervals.
For each positive integer $n$, let $f_n:U_n\to V_n$ be defined by $$f_n(x)=\frac{2x+a_n+b_n}{4}$$ and let $f:\mathbb{R}\to X$ be defined by $$ f(x)= \begin{cases} x&\text{if}\;x\in K\\[4pt] f_n(x)&\text{if}\;x\in U_n\\[4pt] \end{cases} $$ Then it's easily seen that $f$ is an order isomorphism from $\mathbb{R}$ onto $X$.