Is there other homomorphisms from $\mathbb{Z}_q^*$ to $\mathbb{Z}_{pq}^*$? For given two distinct primes p and q, is there other homomorphisms from the multiplicative group $\mathbb{Z}_q^*$ to the multiplicative group $\mathbb{Z}_{pq}^*$, except the following two maps:


*

*$h(r\bmod q)=1$ for all $r\in\mathbb{Z}_q^*$;

*$h(r\bmod q)=r^{p-1} \bmod pq$ for all $r\in\mathbb{Z}_q^*$.

 A: Assuming that we know a generator $g_1$ (resp. $g_2$) for $\mathbb{Z}_p^*$ (resp. $\mathbb{Z}_p^*$) we can describe all the homomorphisms as follows. By the Chinese Remainder Theorem we know that
$$
\mathbb{Z}_p^*\times\mathbb{Z}_q^*\cong\mathbb{Z}_{pq}^*.
$$
Let $d=\gcd(p-1,q-1)$, and let $f:\mathbb{Z}_p^*\to\mathbb{Z}_p^*\times\mathbb{Z}_q^*$ be an arbitrary homomorphism. If we know the image $f(g_1)$, we have fully described $f$, as then $f(g_1^i)=f(g_1)^i$ for all integers $i$, and all the elements of $\mathbb{Z}_p^*$ are of the form $g_1^i$ for some integer $i$.
The constraint is that $f$ has to respect the relation $g_1^{p-1}=1$, so we need to have $f(g_1)^{p-1}=(1,1)$. If $f(g_1)=(a,b)$, this translates to the requirement $b^{p-1}=1$, as the other equation $a^{p-1}=1$ is automatic.
Here $b=g_2^j$ for some $j, 0\le j<q-1,$ so $b^{p-1}=1$ holds, if and only if $(q-1)\mid j(p-1)$. Here the help from factor $p-1$ is exactly the gcd, so this holds, iff $j$ is divisible by $(q-1)/d$. So we have exactly $d$ choices for $j$, namely $j=k(q-1)/d$ for some $k$ in the range $0\le k<d$. Let us write
$z=g_2^{(q-1)/d}$. If $j=k(q-1)/d$, we then have 
$$b=(g_2)^{k(q-1)/d}=z^k.$$
Putting this altogether we get that
$$
f(g_1^i)=(a^i,z^{ki}).
$$
There are $p-1$ choices for $a$ and $d$ choices for $k$, so altogether we have $d(p-1)$ distinct homomorphisms.
If you want to take into account the isomoprhism given by CRT, then you would need to find the images of the pairs $(g_1,1)$ and $(1,z)$ in $\mathbb{Z}_{pq}^*$. Alternatively you can just try and find the $d(p-1)$ elements $w=w_j, j=1,2,\ldots,d(p-1),$ of $\mathbb{Z}_{pq}^*$ that satisfy the equation $w^{p-1}=1$. Then you get all the homomorphisms from the recipe
$$
f(g_1^i)=w_j^i.
$$
