This is from Dummit and Foote Chapter 6.2 Exercise 29 and there is a hint to work in the permutation representation of degree $819$.
In Exercise 28 we reduced to:
$n_3=2863$
$n_7=47853$
$n_{13}=25767$
$n_{409}=819$
I found a proof here: https://www.slader.com/textbook/9780471433347-abstract-algebra-third-edition/214/exercises/29/
What I don't understand is, why can't we have besides an orbit of $1$ and an orbit of order $409$ a second orbit of order $409$. This would neatly sum up to $819$, the number of Sylow $409$-subgroups.
It's easy to see that NG(P) is not an abelian group (otherwise the normalizer of sylow 3 subgroup would have prime factor 409). Then n3 of NG(P) is 409.
Suppose Q is another sylow 409 subgroup, denote NG(P)=U, NG(Q)=V, then U intersects with V is a group of order 1 or 3 (it can't be 409 since otherwise they would share the same sylow 409 group).
That means there exists a sylow 3 subgroup in U\V, acting on Q as an orbit of length 3 by conjugation. That is, U acts on Q by conjugation results in an orbit of length 3*409, overflowing n409.