Is the structure $(\mathbb{R},<,0,1)$ rigid?

Question

Is there a non-trivial automorphism of the structure, $(\mathbb{R},<,0,1)$? I suspect there is, but cannot exhibit an example of one.

Answer 1

More generally, note that $\mathcal{R}_0=(\mathbb{R};<)$ is a highly homogeneous linear order: given finite sequences of reals $$a_1<a_2<...<a_n\quad\mbox{and}\quad b_1<b_2<...<b_n$$ there is an automorphism of $\mathcal{R}_0$ sending $a_i$ to $b_i$. This is just because all nonempty open intervals (including unbounded ones) are order-isomorphic, so we can just "pair up" the various sections $(-\infty, a_1)$ with $(-\infty, b_1)$, $(a_1, a_2)$ with $(b_1, b_2)$, etc.

This means that even after fixing finitely many points, we still have lots of automorphisms; e.g. to get a nontrivial automorphism of the expansion $(\mathcal{R}_0; 0,1)$ just use the above homogeneity fact with $a_1=0, a_2=1, a_3=2$ and $b_1=0, b_2=1, b_3=3$.

Answer 2

$\sqrt[3]\bullet$ seems about right.