This seems to now be Exercise 80 in the latest draft (April 20th).

\\(F\\) does indeed have a right adjoint \\(R : \mathbb{N} \to \mathbb{N}\\); it is defined by \\(R(x) = 3(x + 1) - 1\\). (One can alternately write \\(3x + 2\\), but this version changes less as you change the modulus.)

\\(R\\) does _not_ have a right adjoint of its own. Suppose it did, and call it \\(R'\\). Then \\(0 \le R'(0)\\), since 0 is the minimum of \\(\mathbb{N}\\). So we expect \\(R(0) \le 0\\); however, \\(R(0) = 3(0 + 1) - 1 = 2\\), which is strictly greater than \\(0)\\). Therefore, \\(R\\) cannot have a right adjoint.