Let’s define asymmetric fraction of a finite group $G$ as the number $af(G) = \frac{|\{(g, a) \in G \times Aut(G)| a(g) = g\}|}{|G||Aut(G)|}$. Equivalently it can be defined as $P(A(X) = X)$, where $A$ and $X$ are independent uniformly distributed random elements of $Aut(G)$ and $G$ respectively.

Is it true, that $\forall \epsilon > 0 \exists N \in \mathbb{N} \forall G ((\lvert\,G\,\rvert > n) \to (af(G) < \epsilon))$?

I know, that $af(C_{p^n}) = \dfrac{p^n + \Sigma_{i = 1}^n  p^ip^{n - 1 - i}(p - 1)}{p^{2n - 1}(p - 1)} = \dfrac{(np - n + 1)}{p^n(p - 1)}$ and, that $af(G) \leq \frac{1}{2} + \dfrac{|\{g \in G| \forall a \in Aut(G) \text{ } a(g) = g\}|}{2|G|}$. However this is clearly not enough to prove the statement. 

[This question was already asked by me on MSE, but received no answers][1]

[1]:https://math.stackexchange.com/q/3328203/407165