# When Aut(M) preserves a linear order?

I have a general-type question: Suppose $M$ is a countable structure that is ultrahomogeneous, i.e. every (partial) isomorphism between finitely generated substructures of $M$ extends to an automorphism of $M$. Also, let $Aut(M)$ be the group of automorphisms of $M$ equipped with the pointwise convergence topology.

I want to know necessary and/or sufficient conditions under which $Aut(M)$ preserves a linear ordering on $\mathbb{N}$, in the sense that $Aut(M)$ preserves $(\mathbb{N},\prec)$ iff for every $n,m\in\mathbb{N}$ and for every $g\in Aut(M)$, $$n\prec m\Leftrightarrow g(n)\prec g(m).$$

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Conditions on $M$ or conditions on $\mathrm{Aut}(M)$? A necessary condition is that $\mathrm{Aut}(M)$ is torsion-free. –  Emil Jeřábek Oct 31 '11 at 17:06
Are you assuming that there already is a definable ordering in your setup and asking for conditions to have an infinite ascending chain, or are you asking about situations where you can prove that there is such a definable ordering? –  Scott McKuen Oct 31 '11 at 17:19
If you're interested in the usual ordering on $\mathbb N$, not a general one, then that happens if and only if, for each element, there is a formula that uniquely picks it out. –  Will Sawin Oct 31 '11 at 18:13
@Will: I don’t think so. Let $M$ be $\mathbb Z$ (as a set) with all unary predicates of the form $x\equiv a\pmod b$ for natural numbers $0\le a< b$. Then every two elements of $M$ are distinguishable by an atomic formula, hence $M$ is ultrahomogeneous with a trivial automorphism group (in particular, it preserves whatever you please), but no element of $M$ is definable (because every reduct of $M$ to a finite language has a nontrivial automorphism, namely a suitable shift). –  Emil Jeřábek Oct 31 '11 at 18:39
Even more generally, if $g\in\mathrm{Aut}(M)$ and $a\in M$ are such that $g^n(a)=a$ for some $n>0$, then $g(a)=a$. If not, then WLOG $a\prec g(a)$. Since $g$ preserves $\prec$, we have $g^k(a)\prec g^{k+1}(a)$ for every $k< n$, which gives $a\prec g^n(a)=a$ by transitivity, a contradiction. –  Emil Jeřábek Nov 2 '11 at 10:47