Let $p$ be an odd prime number. Let $K/\mathbb Q_p$ be a finite extension, with $K$ having residue field $\mathbb F_q$ of order $q$ some power of $p$. Let $\zeta\in\mu_{q-1}\subset K$ be a $(q-1)$st root of unity in $K$. Let $\tau\in\mathrm{Aut}(K)$ be an automorphism of $K$ of order $r$, with $r$ some power of $p$.
Assume that the $\tau$-norm of $\zeta$ is $1$, that is, $\zeta\cdot\tau(\zeta)\cdots\tau^{r-1}(\zeta)=1$. Does it follow that $\zeta=1$?
I believe I have found a counter-example. Let me set the scene.
Let $K=\mathbb{Q}_p(\zeta)$, with $\zeta$ a primitive $(q-1)$'th rooth of unity, and let $\tau\in \text{Aut}(K)$. Let's assume $\tau$ fixes $\mathbb{Q}_p$. Then $\tau$ is completely determined by $\tau(\zeta)$, and this is clearly again a primitive $(q-1)$'th root of unity, say $\tau(\zeta)=\zeta^k$ for some $0< k<q-1$, $\gcd(k,q-1)=1$. Note also that $\tau$ has the same action on all the $(q-1)$'th roots of unity.
First of all, note that the order of $\tau$ is then the multiplicative order of $k$ in $(\mathbb{Z}/(q-1)\mathbb{Z})^\times$, which divides $\varphi(q-1)$.
Let us take a look at the $\tau$-norm of an arbitrary $q-1$ root of unity $\zeta_i$, noting that $\tau(\zeta_i^m)=\zeta_i^{mk}$: $$ \zeta_i\cdot\tau(\zeta_i)\cdots\tau^{r-1}(\zeta_i) = \zeta_i\cdot\zeta_i^k\cdots\zeta_i^{k^{\text{ord}(k)-1}} = \zeta_i^{\frac{k^{\text{ord}(k)}-1}{k-1}}.$$
Note crucially that $k^{\text{ord}(k)}-1=0$ mod $q-1$ by definition of the order.
Thus, if we can simply pick $k<q-1$ such that $\gcd(k,q-1)=1$, $(k-1)\mid (q-1)$ and $p|\text{ord}(k)$, then we could choose $\zeta_i$ to be a $\frac{q-1}{k-1}$'th root of unity which is not $1$, and it would have norm $1$, a counter-example.
This is indeed possible, and you can find many examples easily with a computer. The smallest example is $p=3$, $q=27$, $k=3$. It is obvious that $k$ has order $3$ in $(\mathbb{Z}/26\mathbb{Z})^\times$, so let $\tau(\zeta_{26})=\zeta_{26}^3$, an automorphism of $K=\mathbb{Q}(\zeta_{26})$. Picking $\zeta_i$ to be any $13$'th root of unity among the group of $26$'th roots of unity, we see that it has norm $$\zeta_i^{\frac{3^3-1}{3-1}} = \zeta_i^{13} = 1$$ by our above computations. This completes the counter-example.
For completeness, below are more counter-examples. Here $q=p^r$. This is an exhaustive list for $p\leq 13$ and $r\leq 5$.
$p=3$, $r=3$, $k=3$, $\text{ord}(k)=3$.
$p=5$, $r=3$, $k=3$, $\text{ord}(k)=30$.
$p=5$, $r=5$, $k=3$, $\text{ord}(k)=70$.
$p=5$, $r=5$, $k=5$, $\text{ord}(k)=5$.
$p=13$, $r=5$, $k=5$, $\text{ord}(k)=3094$.
$p=13$, $r=5$, $k=7$, $\text{ord}(k)=3094$.
There are many examples for every $p=3,5,7,13$ with $r=6$. Interestingly, no examples for $p=11$ with $r\leq 6$.