Prove that $G$ is abelian if there exists $m\in \mathbb{N}$, so that $f(x)=x^m$, $f \in Aut(G)$ Let $G$ be a finite group with the following property: for any automorphism $f \in G$, there exists $m\in \mathbb{N}$ such that $f(x)=x^{m},\forall x\in G$.
Prove that $G$ is abelian.
Let $s,t \in G$, with $ord(s)|ord(t)$.
We know that $f:G\rightarrow G, f(x)=txt^{-1} \in Aut(G)$. Therefore $f(x)=x^m, m \in \mathbb{N}, m \geq 1.$
Then $t=t\cdot t\cdot t^{-1}=f(t)=t^m$, hence $t^{m-1}=1$, following that $ord(t)|m-1.$
Obviously, $ord(s)|m-1$, so $s^{m-1}=1$, which means $s=s^m=f(s)=tst^{-1}.$
Multiplying with $t$ on the right side, we obtain $st=ts$.
So I proved that $ts=st$, for any $s,t \in G$, with $ord(s)|ord(t)$.
But I got stuck here, because I don't know how to prove that any two elements of $G$ commute, regardless of their order.
 A: Let $p$ be a prime dividing $|G|$.  Let $H_p$ be the set of all elements in $G$ whose order is a power of $p$.  This is a normal subgroup; clearly it is fixed by all automorphisms, so we need to check it's a subgroup.  If $x,y\in H_p$, then we can assume $|x|$ divides $|y|$ (remember, they each have order a power of $p$). By what you've proved, they commute, and so $|xy|$  divides $|y|$; this means $xy\in H_p$, so it's a subgroup.  Note we've also proved $H_p$ is abelian.
Do this for all primes dividing $|G|$.  Clearly, $H_p\cap H_q$ is trivial when $p\neq q$. Thus they all commute, and so $\prod H_p$ is a normal abelian subgroup of $G$. We'll be done if we can show this is all of $G$.
Let $g\in G$ such that $|g|=\prod_i p_i^{\alpha_i}$. We can induct on $i$ to show that $g$ is in $\prod H_p$. The case $i=1$ is obvious.
For the inductive case, write $|g|=p^\alpha r$, where $r$ is "the rest" of the factors. Since $p^\alpha$ and $r$ are coprime, we can write $mp^\alpha+nr=1$.  Then $g^{mp^\alpha}g^{nr}=g$, and $g^{nr}$ has order a power of $p$.  Thus $g^{nr}\in H_p$, and by the induction hypothesis, $g^{mp^\alpha}\in\prod H_p$. Thus
$$ g=g^{mp^\alpha}g^{nr}\in \prod H_p$$
EDIT:
I should mention that once you know $G$ is abelian, it is pretty easy to prove it must be cyclic.
A: I have made corrections in view ofthe mistake pointed out by Max and the observation of Derek Holt. Thanks Max and Derek!
Here is my modified argrument. Let $\vert G\vert=p_1^{n_1}p_2^{n_2}\ldots p_k^{n_k}$.  For each $g\in G$, there is a smallest $x\in\mathbb{N}$ such that $tgt^{-1}=g^x$ for all $g\in G$.  Let us denote this $x$ by $\psi(t)$.  Then, \begin{equation}g^{\psi(s)\psi(t)}=g^{\psi(st)}\forall g\in G.\ \ \ (*)
\end{equation}   
From (*) we have $o(g)\mid (\psi(s)\psi(t)-\psi(st))$ for all $g\in G$ for any $s$, $t\in G$. Let
$m=lcm\{o(g)\vert g\in G\}$, i.e. the exponent of $G$. Then, it follows that $\psi(ts)\equiv \psi(t)\psi(s)\pmod{m}$. If $o(t)=k$, $1=\psi(t^k)\equiv \left(\psi(t)\right)^k\pmod{m}$, so $\psi(t)$ is coprime to $m$. Consider the map $\lambda\colon g\to\left(\frac{\mathbb{Z}}{m\mathbb{Z}}\right)^*$ given by $t\mapsto \{\psi(t)\}$, where $\{\psi(t)\}$ is the residue class of $\psi(t)\bmod{m}$.  So, $\lambda(G)\subset \left(\frac{\mathbb{Z}}{m\mathbb{Z}}\right)^*$.  From ( * ) we actually have group homomorphism $\lambda \colon G\to \left(\frac{\mathbb{Z}}{m\mathbb{Z}}\right)^*$.The kernel of $\lambda$  is $Z(G)$. 
Again,  $G$ is the direct product of its $p$-Sylow subgroups since the $p$-Sylow subgroups are normal as pointed out by Derek Holt.  Let $H_p$ be the $p$-Sylow subgroup of $G$. For any $t\in H_p$, there is an inner automorphism $t\mapsto tgt^{-1}$, so, we still have the map $\lambda$ for this case. The image of $\lambda$ restricted to $H_p$ is in $\left(\frac{\mathbb{Z}}{p^r\mathbb{Z}}\right)^*$ for some $r$. The kernel of the map is $Z\left (H_p\right)$ and $H_p/Z\left(H_p\right)$ is isomorphic to a subgroup of $\left(\frac{\mathbb{Z}}{p^r\mathbb{Z}}\right)^*$. So, if $p$ is odd, $H_p/Z \left (H_p\right)$ is cyclic. It follows that $H_p$ is abelian.  The argument doesn't work for $2$.
