Conjugacy classes in virtually nilpotent groups
It is not possible for an infinite order element $g$ of a virtually nilpotent group to be conjugate to $g^n$ for $n>1$.
Suppose $t \in G$ with $t^{-1}gt = g^n$. Then the subgroup $H = \langle t,g \rangle$ of $G$ is isomorphic to a quotient of the solvable Baumslag-Solitar group $X = = \langle g,t \mid t^{-1}gt=g^n \rangle$.
Now $X$ has the abelian normal subgroup $Y := \langle g^G \rangle$, which is isomorphic to the additive group of rational numbers that can be written as $a/n^k$ for some $a,k \in {\mathbb Z}$. It is not hard to see that any nontrivial quotient of $Y$ is a torsion group, and also that no nontrivial normal subgroup of $X$ can intersect $Y$ trivially. So, since we are assuming that $g$ has infinite order, we have $H \cong X$.
But $X$ is not virtually nilpotent. To see that, show that no subgroup of $X$ of finite index can have trivial centre.
For a virtually abelian group $G$, the issue can be settled with some Euclidean geometry. The general structural theorem for virtually abelian groups is as follows (this is a summary of the Bieberbach Theorems):
- $G$ has a unique maximal finite normal subgroup $N < G$,
- The quotient $G / N$ is a Euclidean crystallographic group, meaning that is has a faithful, discrete, cocompact action (or representation) $G / N \mapsto \text{Isom}(\mathbb{E}^n)$ for some $n \ge 0$.
Here I am using $\mathbb E^n$ to denote Euclidean $n$-space, i.e. $\mathbb R^n$ equipped with the standard metric, and $\text{Isom}(\mathbb E^n)$ is its group of isometries.
Let me denote the composed homomorphism $$f : G \to G/N \to \text{Isom}(\mathbb E^n) $$ Its image is a discrete group, and it follows that an element of $f(G)$ has finite order if and only if it fixes a point.
The key facts regarding order of elements are as follows:
- For each $g \in G$, the translation length $L_g$ of the isometry $f(g) : \mathbb E^n \to \mathbb E^n$ is a conjugacy invariant of $g$.
- $L_g = 0$ $\iff$ $f(g)$ has finite order in $\text{Isom}(\mathbb E^n)$ $\iff$ $g$ has finite order in $G$.
So if $g$ has infinite order then it is a translation with translation length $L_g > 0$, and it follows that $g^n$ is a translation with translation length $n \cdot L_g \ne L_g$. Since translation length is a conjugacy invariant, $g$ and $g^n$ are not conjugate.
I suspect there is also a geometric solution to the virtually nilpotent case, as an alternative to the answer of @DerekHolt. In outline, $G$ still has a unique maximal normal subgroup $N$, and $G/N$ embeds as a lattice in a nilpotent Lie group $\Gamma$, and I believe that one can then proceed using the geometry of the left invariant metric on $\Gamma$.
All finite groups are virtually nilpotent, so consider the quaternion group $Q_8$. Then
$j i j^{-1} = j i (-j) = - i j (-j) = -i = i^{3}$.
So an element is conjugate to its cube, so the answer to your first question is "Yes".