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Is it possible to have a group automorphism of the additive group of the real numbers that fixes a subset of real numbers but is not the identity? The subset might be infinite. I'm thinking of using the fact that real numbers are a vector space over rational numbers, but I'm unsure if this works. I know that an 'ordinary' automorphism such as h(x)=cx won't work.

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It depends on what the subset is, of course. For example, no non-identity automorphism of $\mathbb{R}$ fixes all of $\mathbb{R}$!

But if $A\subseteq \mathbb{R}$ with $|A| < |\mathbb{R}|$, then we can find a non-identity group automorphism of $\mathbb{R}$ which fixes $A$. Your idea is exactly right: View $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and let $B = \text{Span}_\mathbb{Q}(A)$. Then $|B| = |A|$ if $A$ is infinite, and $|B| = \aleph_0$ if $A$ is finite, so $B$ is a proper subset of $\mathbb{R}$. Pick a basis $X$ for $B$, and extend it to a basis $Y$ for $\mathbb{R}$. Then any permutation of $Y$ which fixes all the elements of $X$ induces an vector space automorphism (hence a group automorphism) of $\mathbb{R}$ fixing $B$ (hence fixing $A$).

It's worth noting that there are no non-identity automorphisms of $\mathbb{R}$ as a field (i.e. automorphisms that respect multiplication as well as addition).

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Hint: The vector space idea will work. Choose a basis $H$ that contains $1$, map $1$ to itself, interchange two other basis elements, and map the others to themselves. Extend by $\mathbb{Q}$-linearity.

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