for every $\sigma\in \rm Aut(G)$ and for every $x\in G$, $\sigma(x)=x$ or $\sigma(x)=x^{-1}$ Let $G$ be a group such that for every $\sigma\in \rm Aut(G)$ and for every $x\in G$, $\sigma(x)=x$ or $\sigma(x)=x^{-1}$.
I want to prove that $G$ is solvable.
 A: Note that in such a group, every subgroup is characteristic so $G$ is abelian (a group with all subgroups normal is either abelian or has a direct factor of $Q_8$, but every such latter group has a cyclic subgroup of order 4 not left invariant by an automorphism of order 3).
The torsion subgroup is left invariant, and $G/t(G)$ has the same property, so it cannot be $p$-divisible for any prime $p$.
If $G$ is periodic, then $G$ must be cyclic of order 1,2,3 or 6: its Sylow $p$-subgroups have automorphisms of order $k$ for every $k$ coprime to $p$, but then every such number must be congruent to $\pm1$ mod $p$, and the only such $p$ are $p=2$ and $p=3$. Indeed any element of order larger than $2$ or $3$ would be subject to an automorphism of order other than $2$, so $G$ has exponent dividing $6$, and so is a direct sum of cyclic groups (of orders 2 and 3). If it had more than one direct factor of either order, then it would have a non-characteristic subgroup. Hence $G$ is a subgroup of $C_6$.
What about torsion-free and mixed groups? Must a torsion-free group be rank 1? Are there any mixed examples?
A: WE claim that $G$ is an abelian group. One can see that $Inn(G)=\{id\}$. Thus $\frac{G}{Z(G)}\cong\cong \{id\}$ and so $G=Z(G)$. the claim is proved. Since every abelian group is solvable group, we done.
A: In such a group, every automorphism $\sigma$ is trivial or has order $2$. Hence $\operatorname{Aut}(G)$ is abelian. In particular, $\operatorname{Inn}(G) \cong G / Z(G)$ is abelian, and thus $G$ is solvable.
