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

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

Also, if we restrict the question to abelian case, then the statement will be true. It was proved by Gary Sherman in "What is the Probability an Automorphism Fixes a Group Element?"

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

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