I am working through Aluffi's Algebra Chapter $0$ and I'm not sure how the author intended us to use the conclusion from exercise $4.9$ in $4.10$
Since $p$ and $q$ are distinct prime numbers, they are relatively prime. So using $4.9$, $C_p \times C_q \cong C_{pq} \cong (\mathbb{Z} / pq \mathbb{Z})^*$. However $|(\mathbb{Z} / pq \mathbb{Z})^*| \ne pq$. The order of $(\mathbb{Z} / pq \mathbb{Z})^*$ is $pq-p-q+1 = (p-1)(q-1)$. Since it is assumed that $p$ and $q$ are odd, $gcd(p-1,q-1) \geq 2$ and we can no longer utilize the result of $4.9$
The mistake here is your statement $C_{pq}\cong (\mathbb{Z} / pq \mathbb{Z})^*$, which is not true, because, as you said, the order of these two groups is different, namely $pq\neq (p-1)(q-1)$. Instead you should conclude $$ (\mathbb{Z}/p)^*\times (\mathbb{Z}/q)^*\cong (\mathbb{Z}/(pq))^* $$ from $\mathbb{Z}/p\times \mathbb{Z}/q\cong \mathbb{Z}/(pq)$. Then you have shown that $N=(p-1)(q-1)$.