What are the automorphisms $(\mathbb R, 0, 1, +, \cdot, \leq)\to (\mathbb R, 0, 1, +, \cdot, \leq)$ and how many are there?
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The identity? That would be quite nice. What's the proof? – May 06 '17 at 16:32
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This is a well known theorem: the only field automorphism $f$ of $\mathbb{R}$ is the identity. You don't even need to assume it respects the order, because it's a consequence of being a field automorphism.
Indeed, $f(x^2)=f(x)^2>0$, for every $x\ne0$. Since every positive element is a square, if $x>y$ we have $f(x-y)>0$, so $f(x)>f(y)$.
Then it's easy to prove that $f(m/n)=m/n$, for every rational $m/n$ ($m,n$ integers, $n\ne0$).
Finish by proving that $f$ has to be continuous with respect to the usual metric.
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From the strict monotonicity of $f$ and from $f(q)=q$ for all $q \in \mathbb Q$ we have, for every $x,$ that $\mathbb Q\cap (x,\infty)=\mathbb Q\cap (f(x),\infty) ,$ so $f(x)=x.$ – DanielWainfleet May 06 '17 at 17:11