It is easy to prove that if $E$ is well-ordered, and if $f$ is a strictly increasing map from $E$ to $E$, then, for all $x$ in $E$, $f(x) \ge x$ (just consider the sequence $x$, $f(x)$, $f(f(x))\dots$). But is the converse true, i.e. for any totally ordered set $E$ which is not well-ordered, does it exist a strictly increasing map $f$ from $E$ to $E$ and an element $a$ in $E$ such that $f(a) < a$ ? Even assuming choice, I couldn't find a proof (or a counterexample) ; Cantor-Bendixon (or its generalisation to surreals) seems involved, but it could be a red herring. Any hint?



A characterisation of well-ordering ?
There are dense subsets $X$ of the real line with the usual order (hence not well-ordered) such that the only strictly increasing map from $X$ to itself is the identity. Here's a sketch of the construction. First note that, for any dense set $X$ of reals, an increasing map from $X$ to itself extends to an increasing map on the reals (not necessarily uniquely, beacuse there may be countably many jumps). Such extensions $f$ are determined by their values at the rationals plus some information about jumps; in particular, there are only continuum ($\mathfrak c$) many possibilities. Well-order the set of all such possibilities $f$ in a sequence of length $\mathfrak c$ (the initial ordinal). Build the desired $X$ in $\mathfrak c$ stages, putting one number into $X$ and one into the complement of $X$ at each step, choosing these numbers so as to defeat one possible $f$ at each step. 


