Is there a left-orderable profinite group?

Question

Is there a nontrivial profinite group $G$ with a binary transitive relation $<$ such that

1. $x<y$ implies $x\neq y$, and for any different $x,y \in G$ either $x < y$ or $y < x$ and such that for any $x,y,z \in G$ we have that $x < y$ implies that $zx < zy$ (i.e., $<$ defines a left-invariant strict total order)
2. $\{(x,y):x<y\}$ is open?

Answer 1

There is no such ordering, which is compatible with the profinite topology in the following way: If $x<y$, then there are small neighbourhoods $U, V$ of $x, y$, such that $u<v$ for all $u\in U, v\in V$.

To see this note that If $x>1$, then $x^n>1$ for all $n>0$ and $x^n<1$ for all $n<0$. But in the pro-finite topology, the sequence $x^{n!-1}$ converges to $x^{-1}$, so the set $\{x:x\geq 1\}$ is not closed. But this contradicts our assumption that the ordering is compatible with the topology.