It is easy to prove that if $E$ is wellordered, 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 wellordered, 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) ; CantorBendixon (or its generalisation to surreals) seems involved, but it could be a red herring. Any hint?

There are dense subsets $X$ of the real line with the usual order (hence not wellordered) 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. Wellorder 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. 


In the context of ZF, rather than ZFC, it is consistent that the properties are not equivalent. This is because it is known to be consistent with ZF that there can be an infinite sets of reals that is Dedekind finite, meaning that it has no countably infinite subset. Such a set is linearly ordered by the usual relation on the reals, but it is not wellordered by this relation and indeed, has no wellorder at all, because any such wellorder would allow us to find a countably infinite subset. But meanwhile, there can be no orderpreserving function $f$ with $f(a)\lt a$, since by iterating such a function, we would produce a descending sequence, which would give a countably infinite subset, a contradiction. So these sets are not wellordered by the usual order on the reals, but they have no witnessing function as you request. So the real question, it seems, should be about the ZFC context, for which I am keen to see an answer. 

