Disclaimer: This is not exactly an explanation, but a relevant attempt at understanding conjugates and conjugate classes.
Here is the "conjugation table" of $Q$, where each element in the table is equal to the row header conjugated by the column header, and the end of each row is the conjugacy class of the row header.
The table shows that $1$ and $-1$ form separate conjugacy classes.
\begin{array}{|r|rr|rr|rr|rr|c|}
\hline & 1 & -1 & i & -i & j & -j & k & -k & \text{Conjugacy Class} \\
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \{1\} \\
\hline -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & \{-1\} \\
\hline i & i & i & i & i & -i & -i & -i & -i & \{i,-i\} \\
-i & -i & -i & -i & -i & i & i & i & i & \{i,-i\} \\
\hline j & j & j & -j & -j & j & j & -j & -j & \{j,-j\} \\
-j & -j & -j & j & j & -j & -j & j & j & \{j,-j\} \\
\hline k & k & k & -k & -k & -k & -k & k & k & \{k,-k\} \\
-k & -k & -k & k & k & k & k & -k & -k & \{k,-k\} \\
\hline
\end{array}
Example: $j$ conjugated by $-i$ is $(-i)\cdot j\cdot(-i)^{-1} = -j$.
Might also be noteworthy, to speed up the calculation of conjugations inside $Q$:
- First two rows. $1$ and $-1$ form their own conjugacy class because they commute with all elements in $G$: $$\forall g \,[g\in G \rightarrow [g\cdot g^{-1} = 1 \land g\cdot 1\cdot g^{-1} = 1 \land g \cdot (-1)\cdot g^{-1} = -1]].$$
- First two columns. For the same reason, every element conjugated by $1$ or $-1$ is itself: $$\forall g \,[g\in G \rightarrow 1\cdot g\cdot 1^{-1} = (-1)\cdot g\cdot (-1)^{-1} = g].$$
- Every two rows grouped. Negate a single factor, and you negate the entire product. Therefore, every such two rows are negation of one another: $$\forall g\forall h \,[(g\in G \land h\in G) \rightarrow -(g\cdot h\cdot g^{-1}) = g\cdot (-h)\cdot g^{-1}].$$
- Every two columns grouped. Two negations make a positive in products, therefore every such two columns are equal, e.g. $$\forall g\forall h \,[(g\in G \land h\in G) \rightarrow g\cdot h\cdot g^{-1} = (-g)\cdot h\cdot (-g)^{-1}].$$
Therefore, one only needs to figure out the first two rows, first two columns, and each of the 9 cases where $i,j,k$ gets conjugated by $i,j,k$, to populate the entire conjugation table and obtain conjugacy classes.
In conclusion, there clearly are five conjugacy classes: $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$ and $\{\pm k\}$.
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.
Best Answer
Hint 1: If $b\in Z(G)$ and $x\in G$, what is $xbx^{-1}$?
Hint 2: In the definition of $Z(G)$, write the condition "$ba=ab$" in a way that makes one side a conjugate of $b$.
Otherwise, what you say isn't meaningful. Every element of a group has a conjugate (namely itself), regardless of whether it is central or not. A subgroup being a union of conjugacy classes is a generally non-trivial thing (though here it is relatively simple). They're called normal subgroups, and not all subgroups are normal in general.