Q:

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.

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

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

A:

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. 

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.

