Let $G$ be a group with identity element $e.$ Suppose that for any two elements $a,b\in G-\{e\}$ there exists an automorphism $f$ defined by $f(a)=b.$
a) Prove that all elements of $G-\{e\}$ have the same order.
b) If $G$ is finite, prove that it is abelian.
I attempt to solve part (a) as follows: Let $\vert a \vert=m$ and $\vert b \vert=\vert f(a)\vert=n,$ for $m,n\in Z^+$
Then $a^m=e$ and $b^n=f(a)^n=e.$
But, $b^m=f(a)^m=f(a^m)=f(e)=e.$ So that $n/m.$
Also, $f(a^n)=f(a)^n=e=f(e),$ so that $a^n=e.$ Thus, $m/n.$ Hence $m=n.$
But, I am not sure whether it is correct or not.
I don't know how to solve part (b). I need a help on how to solve it. Thanks in advance!
Your proof for (a) is correct. For (b) a hint: under your condition $Z(G)$, the center of $G$ is the full group and it is thus abelian, unless $Z(G)$ is trivial. By (a), the transitivity implies that all elements of $G$ except $1$ have the same order. Show that then this order has to be prime and $G$ is a non-trivial $p$-group. But then, it is folklore that the center of a non-trivial $p$-group is non-trivial. So $G$ is a finite abelian $p$-group in which all elements (except $1$) have order $p$. That is, it is an elementary abelian $p$-group and isomorphic to a finite direct product of copies of $C_p$. (If there are $n$ copies, $Aut(G)=GL(n,p)$ and this group works indeed transitively on the vector space $C_p \times \cdots \times C_p$.)