Recall that two elements $h_1,h_2$ of a finite group $G$ are called conjugate when $h_2 = gh_1 g^{-1}$ for some $g \in G$, and algebraic-conjugate when $h_2 = gh_1^a g^{-1}$ for some $a \in (\mathbb{Z}/\mathrm{order}(g))^\times$; equivalently when the cyclic subgroups $\langle h_1\rangle$ and $\langle h_2 \rangle$ are conjugate. In general, algebraic-conjugacy is a much coarser relationship than conjugacy. But sometimes they are equivalent: in the symmetric group, for example, both conjugacy and algebraic-conjugacy are simply the cycle structure of the element.
How common is it for algebraic-conjugacy to equal conjugacy?
One reason why you might ask this is the following. Suppose you have a representation $V$ of $G$. Then its character is, of course, a class function: $\mathrm{tr}_V(g)$ depends only on the conjugacy class of $g \in G$. But if $V$ happens to be defined over $\mathbb{Q}$, then $\mathrm{tr}_V(g)$ in fact depends only on the algebraic conjugacy class of $g$. In particular, if every representation of $G$ is defined over $\mathbb{Q}$, then by character theory, algebraic-conjugate elements are automatically conjugate.
An intermediate relation between conjugacy and algebraic conjugacy is what I'll call, based on its character-theory interpretation, real conjugacy: I'll say that $h_1$ and $h_2$ are real-conjugate if $h_2 = g h_1^{\pm 1}g^{-1}$ for some $g$ and some sign.
How common is it for algebraic-conjugacy to equal real-conjugacy?