Is this a characterization of well-orders?
While grading some papers and thinking about a question related to well-orders (in particular, pointing a mistake in a solution), I came to think of a reasonable characterization for well-orders. I can immediately see it's true for countable orders, but not for uncountable orders.
Definition. Let $\cal L$ be a first-order language, and $\cal M$ an $\cal L$-structure. We say that $\cal M$ is rigid if $\text{Aut}(\cal M)=\{\rm id\}$.
Conjecture. Let $\cal L=\{<\}$. Then $\mathcal M=\langle M,<\rangle$ is rigid if and only if it is a well-order or its reverse order is a well-order.
One direction is trivial. Well-orders are rigid (and therefore their reversed orders are rigid as well). In the other direction, if $\cal M$ is countable, then it is easy. Suppose it's not a well-order.
- If $\cal M$ contains a convex copy of $\Bbb Z$ then fix everything else and shift that copy by $1$.
- Otherwise $\cal M$ contains a convex copy of $\Bbb Q$ and then we can shift that copy by $1$ (or multiply it by $\frac12$, whatever floats your boat).
The problem is that for uncountable order types the plot thickens and they might be $\kappa$-dense, so neither of the arguments would work (and there is no additive structure - that I know of - that we can exploit like in the countable case).
Question: Does the conjecture hold, or is there some intricate counterexample?
The conjecture does not hold. For a simple counterexample, consider $\omega+\omega^*$.
In fact, there are rather dramatic counterexamples: In this answer, Brian M. Scott refers to
B. M. Scott. A characterization of well-orders, Fundamenta Mathematicæ, 111 (1), 71-76. MR0607921 (82i:06001),
where he proves that for every infinite $\kappa$ with $2^{<\kappa}=\kappa$ there is a rigid dense linear order of size $2^\kappa$. Moreover, this order does not even admit order-preserving injections into itself other than the identity.
For $\kappa=\omega$, this is a result of Dushnik and Miller, who even built such a set as a dense subset of the reals. This is Theorem 9.1 of
Joseph G. Rosenstein. Linear orderings, Pure and Applied Mathematics, 98. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1982. MR0662564 (84m:06001).
For an additional example, in his answer, Brian also mentions the set $$ X=\{(n,\alpha)\mid n\in\omega\,\&\,\alpha\in\omega_n\},$$ ordered by $(m,\alpha)\le(n,\beta)$ iff $m>n$ (as natural numbers), or else $m=n$ and $\alpha\le \beta$ (as ordinals). This order is rigid, and neither a well-order, nor a reverse well-order.
In his paper, Brian also characterizes well-orders: Say that a linear order $(X,<)$ is cushioned iff every order preserving injection $f:X\to X$ satisfies $x\le f(x)$ for all $x$.
Theorem. A linear order $(X,<)$ is a well-order iff it is cushioned and scattered.
There is a related notion of rigidity, where we require $(X,<)$ to be rigid iff it admits no isomorphism with a proper initial segment of itself. This notion has been studied by John L. Hickman, see
J. L. Hickman. Rigidity in order-types, J. Austral. Math. Soc. Ser. A, 24 (2), (1977), 203–215. MR0480217 (58 #397).