Suppose $k \in \mathbb{Z}$. Can one get the classification of the torsion groups in which every element is conjugate to its $k$-th power?
I was able to solve this problem $\forall k \neq 1$
Case 1:
If $k = 0$, then the only such group is the trivial one.
Proof:
The group identity is conjugate only with itself.
Case 2:
If $k = 1$, then all groups satisfy this property.
Proof:
Every group element is conjugate with itself.
Case 3:
If $k > 1$, then the only such group is the trivial one.
Proof:
Suppose a non-trivial torsion group $G$ satisfies this condition. We can choose a non-trivial element $a$ with minimal possible order. This elements will be some prime $p$. If $p|k$, then $a^k = e$ is not conjugate with $a$. And if $p$ does not divide $k$, then $\exists g \in \mathbb{G}$, such that $gag^{-1} = a^k$, and conjugation by that $g$ induces an automorphism on $\langle a \rangle_p$. But $|Aut(\langle a \rangle_p)| = p - 1 < p$, thus $\langle g \rangle$ contains a non-trivial element with order less than the one of $a$. Contradiction.
Case 4:
If $k < -1$, then the only such group is the trivial one.
Proof:
The proof is exactly the same as for case 3.
My question however is:
What to do with $k = -1$?
For $k = -1$ I know that the following groups satisfy those conditions:
1) the trivial group
2) groups of exponent $2$
3) dihedral groups
4) symmetric groups
However, I do not know, whether there is anything else, and if there is, then how to prove it.