Is the structure $(\mathbb{R}, +, *)$ rigid?

Question

Does the structure $(\mathbb{R}, + ,*)$ have only the trivial automorphism?

Answer 1

Suppose $\phi\in\text{Aut}(\mathbb{R})$. $$\phi(1)^2 = \phi(1) \implies \phi(1) = 1 $$ since $\phi$ is an automorphism. It's easy to see that for $x\in\mathbb{Q}$, $\phi(x) = x.$ For $x> 0$ $$\phi(x) = \phi(\sqrt{x})^2> 0. $$ So $$ x>y \implies \phi(x)>\phi(y).$$ If $x$ is irrational, there exists $2$ sequences, one increasing $x_n$, and one decreasing $y_n$, both are sequences of rational numbers, and converge to $x$. $$y_n = \phi(y_n)\geq \phi(x)\geq \phi(x_n) = x_n $$ By taking limit, $\phi(x) =x$ for all $x$.