$n$th power map is an automorphism implies abelian group? 
If $G$ is a finite group and $\phi(x) = x^n$ is an automorphism of $G$ does this imply $G$ is abelian?

I've been reading this page.
Def: A group $G$ is said to be $n$-abelian if $(ab)^n=a^nb^n$ for every $a,b\in G$.
which convinces me my question must be true....but I just want to make sure I'm not misinterpreting anything.
If yes, how would I go about proving this. i.e: showing $xy=yx$ -Thank u
I've seen the proof for $\phi(x) =x^2$. For arbitrary $n$ is this the same?
 A: It is not true for arbitrary $n$,
Being homomorphism only give us that $(xy)^n=x^ny^n$ and it is not enough for arbitrary $n$.
here is required condition.(if it is true for three consecutive $n$, you can conclude $G$ is abelian)
Note: But you can show that it is true for these numbers $n=-1,1,2$
A: Not necessarily true for endomorphisms. For example, there are nonabelian groups where every element has order $3$; so $x\mapsto x^3=1$ is the trivial map, so trivially a group endomorphism. 
A: Let $G$ be any finite non-abelian group and consider its order $|G|$. Then the map $x\mapsto x^{|G|+1}$ is simply the identity, so it is trivially an automorphism.
For the result you want, there must be some condition between $n$ and $|G|$, specifically $\operatorname{mcd}(|G|,n(n-1))=1$. See groupprops.
A: Take any non-abelian group $G$ of order $n$. Then $G$ is $(n+1)$-abelian, since the map $g\mapsto g^{n+1}$ is the identity (hence, an automorphism).  For a specific example, the symmetric group of degree $3$ is $7$-abelian.
