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.