set theory - Automorphisms on $(mathbb R,+)$ and the Axiom of Choice

We know that the algebraic automorphisms of the real numbers under addition is not in $\text{1:1}$ correpondence with $\mathbb R \setminus \{0\}$; see here.

The argument uses the AOC.

Suppose we drop the AOC from $\text{ZFC}$ replacing it with

Axiom (GR):

The injective mapping

$\quad \Phi: \mathbb R \setminus \{0\} \to \text{AutomorphismGroup(} \mathbb R ,+ \text{)}$

is surjective.

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Has this $\text{ZF+GR}$ been tried and/or does this lead to $1 = 0$?

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Update:

Added descriptive set theory tag after looking over links in Noah's answer.

Answer

It is indeed consistent, and in fact is a consequence of the extremely powerful axiom of determinacy.

Specifically, AD implies that every homomorphism from $(\mathbb{R},+)$ to itself is continuous, and in particular of the form $a\mapsto ar$ for some $r\in\mathbb{R}$. See here for some discussion of how nasty any other endomorphism would have to be; AD rules out such sets (e.g. implies that every set of reals is measurable).

Of course, as Asaf observes below, AD is truly massive overkill (like, nuking a mosquito); I'm mentioning it because AD is a natural alternative to AC which you may independently want to know about.

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Now AD isn't actually cheap: the theory ZF+AD proves the consistency of ZF, that is, the axiom determinacy is of high consistency strength. We can prove the consistency of ZF+GR relative to ZF alone; however, this is a bit more technical.