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I know the classic theorem that $(\mathbb{Q},+)$ cannot be expressed as an automorphism group, i.e. there is no group $G$ such that $\mathrm{Aut}(G)\simeq (\mathbb{Q},+)$.

Theorem A. If $L$ is a locally cyclic group with no element of order $2$, then $L$ cannot be expressed as an automorphism group.

But how about $(\mathbb{R},+)$?

I think the answer might be no, and the proof proceeds by showing that if $\mathrm{Aut}(G) \simeq \mathbb{R}$, then some subgroup or quotient $H$ of $G$ will satisfy $\mathrm{Aut}(G) \simeq \mathbb{Q}$, which contradicts Theorem A.

So how can we construct this $H$? Assuming the axiom of choice, we can say $\mathbb{R} = \mathbb{Q}\oplus B$ for some additive subgroup $B$ of $\mathbb{R}$ (just by picking a $\mathbb{Q}$-basis for $\mathbb{R}$). Now I'm tempted to do some Galois-type thing, where you use a "fixed" subgroup $$H=\mathrm{Fix}(B)=\{g\in G : b(g) = g \text{ for all } b\in B \},$$ and then try to say something about $\mathrm{Aut}(H)$ or $\mathrm{Aut}(G/H)$ (where, for the latter, maybe we take the normal closure of $H$). But I can't complete this line of reasoning. It seems key that $\mathrm{Aut}(G)$ splits as a direct sum --- that seems special.

Am I just totally off-base here? Is there some obvious group $G$ whose automorphism group is $\mathbb{R}$?

Related questions

If I can prove that $\mathbb{R}$ is not an automorphism group, then the same would hold for the isomorphic group $\mathbb{R}_{>0}$ of positive reals under multiplication. But how about $\mathbb{R}^\times \simeq \mathbb{R}_{>0}\times C_2\simeq \mathbb{R}\times C_2$ --- can this be achieved as an automorphism group? If $\mathrm{Aut}(G)$ is a direct product of abelian groups, what can be said about $G$?

A related observation is that if $\mathrm{Aut}(G)$ is abelian, then $G$ is nilpotent of rank $\leq 2$. So there is a sense in which $G$ is "almost" abelian.

I know $\mathbb{Q}^\times$ is the automorphism group of $(\mathbb{Q},+)$, and that $\mathbb{R}^\times$ constitutes the continuous automorphisms of $(\mathbb{R},+)$ ... Then there's multiplicative/additive groups of other rings/fields ...

I can't even resolve this question in the case of $\mathrm{Aut}(G) \simeq \mathbb{Z}\times \mathbb{Z}$. For this one I can show that $G'$ would necessarily by cyclic, but that's about it.

I'm very curious about which types of groups can be achieved as automorphism groups.

A quick observation

If $\mathrm{Aut}(G) \simeq \mathbb{R}$, then $G$ must be infinitely generated (or else its automorphism group is countable) and nilpotent of rank $\leq 2$ (this follows for any group whose automorphism group is abelian).

Ehsaan
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    It's certainly not the case 'generically' that if $H\subset Aut(G)$ then there's some subgroup $G'$ of G with $Aut(G')\equiv H$, even if $Aut(G)=H\times K$; consider the automorphism group of $C_p$ for $p$ prime. I don't see anything in your reasoning offhand that uses any properties of $\mathbb{R}$ that don't mirror to this case. – Steven Stadnicki Nov 27 '20 at 16:08
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    Yes, it would have to be some special property of $\mathbb{R}$ that comes up here. Subgroup or quotient would be OK here --- just as long as we can use the fact that $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ to reduce to the case of $\mathbb{Q}$ (which is a known exercise). – Ehsaan Nov 27 '20 at 16:10
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    See also this post. – Dietrich Burde Nov 27 '20 at 16:12
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    @DietrichBurde thank you --- that's related for sure. I wonder which if there are groups $A$ for which it is true, but difficult, to prove that $A$ cannot be achieved as an automorphism group. – Ehsaan Nov 27 '20 at 16:14
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    @Ehsaan: re: your related questions, it's consistent with ZF that every endomorphism of $\mathbb{R}$ is measurable, hence continuous; if this is true then $\text{Aut}(\mathbb{R}) \cong \mathbb{R}^{\times} \cong \mathbb{R} \times C_2$. (With the axiom of choice $\text{Aut}(\mathbb{R})$ is much larger and in particular nonabelian.) – Qiaochu Yuan Dec 06 '20 at 17:56

  • it's consistent with ZF that every endomorphism of R is measurable

    --- Wow.

     – Ehsaan Dec 07 '20 at 16:48
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    Partial answer: $G$ cannot be abelian. If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism of $G$ of order $\leq 2$. $\mathbb{R}$ has no nonidentity elements of finite order, so $x=x^{-1}$ for all $x\in G$. This implies that $G$ is the underlying abelian group of an $\mathbb{F}^2$-vector space $V$, and we have $\operatorname{Aut}(G)=\operatorname{Aut}_{\mathbb{F}_2}(V)$. We must have $\dim V\geq 2$, but this means $\operatorname{Aut}(G)$ contains $\operatorname{GL}_2(\mathbb{F}_2)$ as a subgroup. This would make $\operatorname{Aut}(G)$ nonabelian, contradiction! – diracdeltafunk Jan 07 '21 at 05:56
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    Thanks, I did already know that! Same idea to prove that $\mathbb{Q}$ is not an automorphism group. – Ehsaan Jan 07 '21 at 21:55

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