Condition for triviality of group morphism $\tau : \mathbb{Z}_p \to\mathrm{Aut}(\mathbb{Z}_q)$ : $\tau_\overline{k}(\overline{n}) = \bar{r}^k \bar{n}$ I'm trying to show the following:

Let $p,q$ be prime numbers, $\tau : \mathbb{Z}_p \to \operatorname{Aut}(\mathbb{Z}_q)$ be a group morphism such that  $$\overline{k} \mapsto \tau_\overline{k}, \quad\tau_\overline{k}(\overline{n}) = \overline{r}^k \overline{n} \quad\forall k,n \in \mathbb{Z}$$ and $r^p \equiv 1 \bmod q$. Then $\tau$ is the trivial morphism if and only if $r \equiv 1 \bmod q$.

For the first implication $(\implies)$ I'm guessing that I have to play around with the facts that $\overline{r}^k \overline{n} = 1$ and $r^p \equiv 1 \mod q$. 
I also know that such a morphism sends the element $\overline{k} \in \mathbb{Z}_p$ to an element of an order that divides $p$, but I'm not quite sure about how to connect these facts.
As always, any ideas are welcome. Thanks in advance!

If $\tau_{\bar k}\bar n=\bar n,\forall\bar n\in\Bbb Z_q$, then $r^k\cong1\pmod q,\forall k\implies r\cong1\pmod q$.
Conversely, if $r=\bar1\in\Bbb Z_q$, then $\tau_{\bar k}\bar n=\bar r^k\bar n=\bar1^k\cdot\bar n=\bar n$ is trivial.
