It is well known that if $(P, \leq)$ is a partial order then $\leq$ can always be extended to a linear order. This is sometimes called Szpilrajn´s theorem although it had been previously proved by Banach, Kuratowski and Tarski.


Now suppose that $f$ is an automorphism of $(P, \leq)$ and we want to extend $\leq$ to a linear order in such a way that $f$ remains an automorphism. Of course, this is not always possible since $f$ could have a finite orbit and automorphisms of linear orders can´t have finite orbits; but I wonder if this is the only obstruction. So let $A$ be the collection of all those $f$´s for which it is possible (for instance $Id\_P \in A$); here are my questions:


**1)** Is it true that If $f$ has no finite orbits then $f \in A$?


**2)** Is $A$ a subgroup of $Aut(P, \leq)$?


and the vaguer


**3)** If the answer to 1 is negative. Can we somehow characterize the elements of $A$?


Perhaps this is all well known and studied but I couldn´t find anything at all in the literature, so references are also appreciated.



Extending a partial order while preserving an automorphism
I couldn't find anything in the literature either, but the answer to the first question is positive. Let $G$ be a group acting on a space $X$. Say that $G$'s action on $X$ has the invariant order-extension property provided that every $G$-invariant partial order on $X$ (i.e., partial order $\le$ such that $x\le y$ iff $gx\le gy$, for each $g\in G$) extends to a $G$-invariant linear order on $X$.


I managed to show:


**Theorem.** The following are equivalent for an abelian group $G$:


1. $G$'s action on $X$ has the invariant order extension property.
2. $X$ has a linear $G$-invariant order.
3. No element of $G$ has an orbit of finite size greater than one.


[Here's](http://arxiv.org/abs/1309.7295) the proof.


The answer to Question 1 follows by letting $G$ be the subgroup of permutations generated by $f$.


(By the way, I assume that your no-finite-orbits condition really means: no finite orbits of size greater than one. Fixed points aren't a problem.)


