An abelian, characteristically simple group is divisible (supposedly). This is Exercise 4.13 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.
The Details:
On page 69 of Roman's book, we have

Definition: Let $G$ be a group. A subgroup $H$ of $G$ is characteristic in $G$ if it is invariant under all automorphisms of $G$.

Write this as $H\sqsubseteq G$.
On page 121 ibid., there is the

Definition: A nontrivial group $G$ is characteristically simple if it has no proper characteristic subgroups.

On page 145, the exercise in question starts out as follows:

An abelian group $A$ (written additively) is divisible if for any $a\in A$ and any positive integer $n$, there is a $b\in A$ for which $nb=a$.

The Question:

Prove that a characteristically simple abelian group $A$ is divisible.

Thoughts:
Let $A$ be a characteristically simple abelian group.
Such a group is classified in the first $\blacktriangleright$ of this answer by @YCor. No justification is given. (I would appreciate a reference to this end, please.)
Let $\varphi \in {\rm Aut}(A)$. Then, for any $B\sqsubseteq A$, either $B$ is trivial or $B=A$, and $\varphi(B)=B$ by definition.
Let $a\in A$. I aim to show that there for all $n\in \Bbb N$, there is some $b\in A$ with $nb=a$.
(I have nothing nontrivial to add.)
Please help :)
 A: Every abelian group $A$ has a canonical collection of characteric subgroups $nA = \{ na : a \in A \}, n \in \mathbb{N}$ given by the $n^{th}$ powers. If $A$ is characteristically simple, these subgroups must each be either the entire group $A$ or the trivial subgroup $0$. So we split into two cases:

*

*If $nA = 0$ for some $n \ge 2$ then we must have $pA = 0$ for some prime divisor $p \mid n$ (otherwise, if $pA = A$ for all $p \mid n$ then we would have $nA = A$), so $A$ is an $\mathbb{F}_p$-vector space. I'll leave it as an exercise to check that every $\mathbb{F}_p$-vector space is characteristically simple (the basic idea is that the automorphism group acts transitively on subspaces of a fixed dimension, so no proper subgroups are characteristic). In particular, the exercise is false as stated.

*If $nA = A$ for all $n$ then $A$ is divisible. This is probably what the author had in mind.

In the divisible case we can keep going with a second source of characteristic subgroups: the $n$-torsion subgroups $\sqrt[n]{A} = \{ a \in A : na = 0 \}$ (this is not standard notation; I think $A[n]$ might be standard, but that notation is already too overloaded for my taste) are characteristic, so again must satisfy either $\sqrt[n]{A} = A$ or $\sqrt[n]{A} = 0$. If $\sqrt[n]{A} = A$ then every element is $n$-torsion so $nA = 0$ and $A$ can't be divisible. So we conclude that $\sqrt[n]{A} = 0$ for every $n$, hence that $A$ is torsion-free. Torsion-free divisible abelian groups are precisely vector spaces over $\mathbb{Q}$ (divisibility allows us to define division and torsion-freeness guarantees that the result is unique), and it's again an exercise to show that $\mathbb{Q}$-vector spaces are characteristically simple.
This proves YCor's classification. He singles out the zero abelian group because it's both the zero $\mathbb{F}_p$-vector space for any $p$ and the zero $\mathbb{Q}$-vector space; it is the unique group which satisfies $A = 0$ so it's an edge case for every case of this argument. In particular it's the unique group which is both torsion-free and torsion!

Note that $nA = \text{im} \left( A \xrightarrow{n} A \right)$ while $\sqrt[n]{A} = \text{ker} \left( A \xrightarrow{n} A \right)$. So their characteristic nature comes from an even stronger property of the multiplication-by-$n$ endomorphism on an abelian group, which is that it's natural (commutes with arbitrary homomorphisms of abelian groups); formally, it's a natural transformation from the identity functor $\text{id}_{\text{Ab}} : \text{Ab} \to \text{Ab}$ to itself. It's a nice exercise to show that every such natural transformation has this form; in other words, $\text{End}(\text{id}_{\text{Ab}}) \cong \mathbb{Z}$ (as rings, even).
This naturality property shows that $nA$ and $\sqrt[n]{A}$ have a property stronger than being characteristic: they are natural subgroups, in that arbitrary homomorphisms $f : A \to B$ induce homomorphisms $nA \to nB$ and $\sqrt[n]{A} \to \sqrt[n]{B}$. Formally, they are both subfunctors of the identity functor, and as subfunctors can can be defined as the kernel resp. the image of multiplication-by-$n$ as a natural transformation.
