Fields that can be ordered in more than one way

Consider $F=\mathbb{Q}(x)$. You can order this field in uncountably many different ways. For instance, for every transcendental $\alpha\in\mathbb{R}$, the isomorphism $F\to\mathbb{Q}(\alpha)\subset\mathbb{R}$ sending $x$ to $\alpha$ induces an ordering of $F$, and this ordering is different for each $\alpha$. Since $F$ has only countably many automorphisms (an automorphism is determined by where it sends $x$), these give uncountably many orderings that are not related by automorphisms.

Here's a smaller example: let $f$ be an irreducible quartic polynomial with four real roots. If $\alpha$ is any root of $f$, then $\mathbb{Q}(\alpha)$ can be ordered in four different ways, corresponding to the four real embeddings of $\alpha$

However, most such polynomials $f$ will have Galois group $A_4$, and $\mathbb{Q}(\alpha)$ will have no automorphisms at all.

After reading the excellent answers by Eric Worfsey and Hurkyl, I realized that I already knew another answer to my own question -- somehow it had escaped me when I posted the question, but I think it is worth posting the answer here.

Let $F=\mathbb{Q}(x)$. Any element of $F$ can be written in the form $$f=\frac{a_n x^n + \cdots +a_0}{b_m x^m + \cdots +b_0}$$ with $a_n, b_m \ne 0$ and where all coefficients are integers. Let $P \subset F$ be defined by the condition $f\in P \iff \frac{a_n}{b_m}> 0$, and define $f \prec g \iff g-f \in P$. Then $(F, \prec)$ is an ordered field.

Now contrast this order with the order on $F$ proposed by Eric Worfsey: in that order, one simply chooses a transcendental real number $\alpha$, maps $x \mapsto \alpha$, and uses the order $<$ inherited from the reals.

These two orders are "really different". The order $<$ obtained by mapping $x \mapsto \alpha$ embeds $\mathbb{Q}(x)$ as an ordered subfield of $\mathbb{R}$, but with respect to the order $\prec$ the field $\mathbb{Q}(x)$ is non-archimedean. Specifically, for all $q \in \mathbb{Q}^+$, we have $q \prec x$ and $0 \prec \frac{1}{x} \prec q$; in other words, $x$ is "infinite" and $\frac{1}{x}$ is "infinitesimal" with respect to the rationals in this order.