Finitely generated group has automorphism mapping between two elements of the same order? For a finitely generated group $G$, is it always the case that if two elements $g_1, g_2$ have the same order then there is an automorphism that sends one to the other (i.e. $\phi$ such that $\phi(g_1)=g_2$)?
I can't seem to prove it, but playing around with several groups, abelian or not (to include $D_3$; a cyclic group of prime order and $\mathbb{Z}_4$), I'm unable to refute this.
How could one go about proving this? (assuming indeed it can be proven).
 A: Take $G=S_3\times\mathbb{Z_2}$. It contains an element of order $2$ which is in the center, and an element of order $2$ not in the center. So clearly there is no automorphism which maps one of these elements to the other one.
A: In $S_4$ there is no automorphism that takes a two cycle like $(12)$ to a product of two cycles like $(12)(34)$.
In general, $S_n$ has only inner automorphisms when $n \ne 2,6$ and those preserve cycle structure, so you can construct lots of similar examples.
A: Automorphisms send conjugacy classes to conjugacy classes, while preserving elements' order. In $S_4$, the only distinct conjugacy classes of elements of the same order are the class of the transpositions and the class of the double transpositions. But the former has size $6$, whereas the latter has size $3$, so no one automorphism of $S_4$ sends transpositions to double transpositions, despite having the elements of both classes order $2$. Yet another affordable case where this argument works, verbatim, is $S_5$.  (Incidentally, this proves also that in these two groups the automorphisms are class-preserving, and hence inner.)
