Consider the symmetric group $S_n$ and, for each $\sigma \in S_n$, let $f(\sigma)$ be the number of fixed points of $\sigma$. Now let $g$ be the permutation such that $g(i)=i+1$ for $i=1,...,n-1$ and $g(n)=1$. We can partition $S_n$ in $(n-1)!$ sets of the form $$\sigma _g = \{\sigma, \sigma \circ g , \sigma \circ g ^2 ,..., \sigma \circ g ^{n-1}\}$$ for all $\sigma \in S_n$ such that $\sigma(1)=1$. Let $$h(\sigma _g) = \max \{f(\pi) \mid \pi \in \sigma _g\}. $$ Show that the average of $h(\sigma _g)$ for all $\sigma \in S_n$ such that $\sigma(1)=1$ is $3$ if $n\rightarrow \infty$.
The only thing I can do is show that the average of $f(\sigma)$ is $1$ if $n\rightarrow \infty$ using the formulas in [1]. I also tried to adapt the approaches in [2], but with no success. At last, I attempted to rewrite the problem in terms of permutation matrices, where $f$ is the trace of the matrix and apply some properties, also without success.
EDIT: so, from Carl Schildkraut's comment and also [3], it seems that this value explodes (although very slowly) as $n\rightarrow \infty$, as it's most likely $$\left( \frac{\log(n)}{\log(\log(n))} \right)$$ asymptotically, because, in each set $\sigma_g$, the distribution of the number of fixed points likely is $n$ Poisson distributions, so the expected maximum, according to [3], should be like that. So this question is most likely wrong and this limit may well be unbounded.
[1] https://en.wikipedia.org/wiki/Rencontres_numbers
