Solution:

Let the original list have $n$ integers with mean $m,$ so its sum is $mn.$ Appending $15$ gives $$(m + 2)(n + 1) = mn + 15 \implies m + 2n = 13.$$

Appending $1$ to that enlarged list gives $$(m + 1)(n + 2) = mn + 16 \implies 2m + n = 14.$$

Solving $m + 2n = 13$ and $2m + n = 14$ yields $m = 5$ and $n = 4.$

Thus, the correct answer is A.