This statement becomes correct when restricted to groups $G$ with trivial Fitting subgroup (ie with $F(G) = 1$), but at present I don't know how to prove it in general. The quantity $af(G)$ you consider is the number of ${\rm Aut}(G)$ orbits on $G$ divided by $|G|$.

When $Z(G) = 1$, $G \cong {\rm Inn}(G)$ embeds isomorphically into ${\rm Aut}(G)$, as a normal subgroup. In that case, your quantity $af(G)$ is certainly at most $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$. The ratio $\frac{k(G)}{|G|}$ is the commuting probability of $G$. In the case that $F(G) = 1,$ Bob Guralnick and I proved that the commuting probability tends to zero as $|G| \to \infty$ (this does not require CFSG, though the convergence rate can be shown to be much faster using CFSG. There is also a more elementary proof of that fact due to P.M. Neumann, of which Guralnick and I were unaware at the time).

This statement becomes correct when restricted to groups $G$ with trivial Fitting subgroup (ie with $F(G) = 1$), but at present I don't know how to prove it in general. The quantity $af(G)$ you consider is the number of ${\rm Aut}(G)$ orbits on $G$ divided by $|G|$.

When $Z(G) = 1$, $G \cong {\rm Inn}(G)$ embeds isomorphically into ${\rm Aut}(G)$, as a normal subgroup. In that case, your quantity $af(G)$ is certainly at most $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$. The ratio $\frac{k(G)}{|G|}$ is the commuting probability of $G$. In the case that $F(G) = 1,$ Bob Guralnick and I proved that the commuting probability tends to zero as $|G| \to \infty$ (this does not require CFSG, though the convergence rate can be shown to be much faster using CFSG. There is also a more elementary proof of that fact due to P.M. Neumann, of which Guralnick and I were unaware at the time).

Later edit: Perhaps it would have been better to say at the beginning that it is true that ${\rm af}(G) \to 0$ as $[G:F(G)] \to \infty$, which is really what the quoted results imply.

This statement becomes correct when restricted to groups $G$ with trivial Fitting subgroup (ie with $F(G) = 1$, but at present I don't know how to prove it in general. The quantity $af(G)$ you consider is the number of ${\rm Aut}(G)$ orbits on $G$ divided by $|G|$.

When $Z(G) = 1$, $G \cong {\rm Inn}(G)$ embeds isomorphically into ${\rm Aut}(G)$, as a normal subgroup. In that case, your quantity $af(G)$ is certainly at most $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$. The ratio $\frac{k(G)}{|G|}$ is the commuting probability of $G$. In the case that $F(G) = 1,$ Bob Guralnick and I proved that the commuting probability tends to zero as $|G| \to \infty$ (this does not require CFSG, though the convergence rate can be shown to be much faster using CFSG. There is also a more elementary proof of that fact due to P.M. Neumann, of which Guralnick and I were unaware at the time).

This statement becomes correct when restricted to groups $G$ with trivial Fitting subgroup (ie with $F(G) = 1$), but at present I don't know how to prove it in general. The quantity $af(G)$ you consider is the number of ${\rm Aut}(G)$ orbits on $G$ divided by $|G|$.

When $Z(G) = 1$, $G \cong {\rm Inn}(G)$ embeds isomorphically into ${\rm Aut}(G)$, as a normal subgroup. In that case, your quantity $af(G)$ is certainly at most $\frac{k(G)}{|G|}$, where $k(G)$ is the number of conjugacy classes of $G$. The ratio $\frac{k(G)}{|G|}$ is the commuting probability of $G$. In the case that $F(G) = 1,$ Bob Guralnick and I proved that the commuting probability tends to zero as $|G| \to \infty$ (this does not require CFSG, though the convergence rate can be shown to be much faster using CFSG. There is also a more elementary proof of that fact due to P.M. Neumann, of which Guralnick and I were unaware at the time).