(This CW takes the ideas presented by Jyrki Lahtonen, JSpecter, Derek Holt, and Steve Dalton, and adjusts the question to something more reasonable.)
Original question 2:
No, n = 77 is another such positive integer.
Part of the solution involved proving that no group had n Sylow p-subgroups for p odd. This is necessary because:
Proposition: If a finite group has n Sylow p-subgroups for odd p, then a finite insoluble group has n Sylow p-subgroups for odd p.
The proof is silly: one just realizes that for every odd prime p there is a finite simple group whose order is coprime to p. For p = 3, one takes a Suzuki group, and for p ≥ 5, one takes an appropriate PSL(2,q).
Original question 1 and a better question 2:
Suppose G has n Sylow p-subgroups, one of them P. Let N = NG(P) be the normalizer of P in G, and let K be the normal core of N in G. Since P is normal in N, N is clearly p-solvable. One has that G/K is p-solvable if and only if G is p-solvable, hence we can assume the action of G on the Sylow p-subgroups is faithful as far as p-solvability is concerned.
This suggests we change "solvable" to "p-solvable" for several reasons:
- The solvable case has a weird "no odd p" requirement, and p = 3 can never be interesting.
- The solvable case for any particular n that works has the Feit–Thompson theorem as a corollary, so the proof must include, at least by reference, fairly difficult ideas.
- The p-solvable case in several instances involved no heavy machinery.
- 2-solvable = solvable by Feit–Thompson, so for p = 2 not much has changed
One thus has the more interesting, but probably much harder question:
What are the pairs (n, p) such that every finite group with n Sylow p-subgroups is p-solvable? (again where we ignore (n, p) such that there are no finite groups with n Sylow p-subgroups)
When n is small (n ≤ 32 for me, n ≤ 63 for some) one has a complete list of possible Sylow actions, and one can just check the list. The result is a much larger list now: (n, p) in
{ (3,2), (4,3), (7,2), (7,3), (11,2), (11,5), (13,2), (13,3), (16,3), (16,5), (19,2), (19,3), (23,2), (23,11), (25,3), (27,13), (29,2), (29,7), (31,2), (31,3), (31,5) }
or n ≥ 33. The original question only had n in { 3 } or n ≥ 33.
When n is medium (n ≤ 2499 for me) one has a complete list of primitive groups, and one can often leverage this to understand Sylow actions. Derek used this to find a larger example, n = 77 that works with the stronger requirement of solvability for all p.
When n = 3 and p = 2, then G/K ≅ S3 is p-solvable, and since K ≤ N is p-solvable, G itself must be p-solvable, using no difficult results. Of course, if n = 3, then p divides n − 1 = 2 by Sylow's theorem, but this seems like an unnecessary complication now that Derek has presented his answer.
When n = 77 and p = 2, then we consult the four primitive groups of degree 77 (all of which have at least 3465 Sylow 2-subgroups) to see that N is contained in a maximal subgroup M such that the indices $[G:M],[M:N]$ are either 7,11 or 11,7. Since N is also a Sylow p-normalizer in M, we get that M/K is either dihedral or affine of degree 7 or 11 (four possibilities, D7, D11, AGL(1,7), AGL(1,11)). In all cases M is p-solvable, so the G-core L of M is p-solvable. The possibilities for G/L are similarly restricted: a subgroup of an affine group in degree 7 or 11. In particular, both G/L and L are p-solvable, so G is p-solvable, and n = 77 is another example.
Of course, p must divide n−1, and the list of primitive groups of degree 77 rules out p = 19 (no intermediate M can exist by Sylow's theorem since neither 7 nor 11 is equivalent to 1 mod 19, but none of the primitive groups actually has 77 Sylow p-subgroups). Thus even when we had not specified p = 2, this still was an example.
When n = 143 and p = 71, one gets that N cannot be contained in M, so the action is primitive, but no primitive group of degree n has n Sylow p-subgroups, so this case does not occur.
When n = 143 and p = 2, the argument is more complicated (the action must be imprimitive, the only insoluble action on 11 or 13 points is by L2(11), but the number of Sylow p-subgroups of a normal section of a group divides the number of Sylow p-subgroups of the whole group, and n2(L2(11)) = 55 does not divide 143), but again is an example.