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

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

I know, that $$\mathrm{af}(C_{p^n}) = \dfrac{p^n + \Sigma_{i = 1}^n  p^ip^{n - 1 - i}(p - 1)}{p^{2n - 1}(p - 1)} = \dfrac{(np - n + 1)}{p^n(p - 1)}$$ and, that $$\mathrm{af}(G) \leq \frac{1}{2} + \dfrac{|\{g \in G\mid \forall a \in \mathrm{Aut}(G), \; a(g) = g\}|}{2|G|}.$$ 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