Let $K/\mathbb{Q}_p$ be the unique unramified quadratic extension of $\mathbb{Q}_p$. Then $K$ contains all $(p^2-1)^\mathrm{st}$ roots of unity. Let $\zeta_{p^2-1}\in K$ be a primitive $(p^2-1)^\mathrm{st}$ root of unity. Since $\zeta_{p^2-1} \in K\setminus \mathbb{Q}_p$, it has a unique conjugate in $K$. I suspect that this conjugate must be $\zeta_{p^2-1}^{-1}$, because that seems like the most natural thing for it to be, but I'm stuck trying to prove it.
My Attempt
One way to show that $\zeta_{p^2-1}^{-1}$ is the conjugate would be to show that $\zeta_{p^2-1}\zeta_{p^2-1}^{-1}$ and $\zeta_{p^2-1} + \zeta_{p^2-1}^{-1}$ are both in $\mathbb{Q}_p$. The former is equal to $1$, so it is clearly in $\mathbb{Q}_p$, but I'm unsure about the latter.
The nontrivial Galois automorphism $\sigma$ of $K$ descends to the Frobenius modulo $p$. So $\sigma(\zeta_{p^2-1}) \equiv \zeta_{p^2-1}^p \bmod p$. Since the $p^2-1$ roots of unity are distinct modulo $p$, this shows that $\sigma(\zeta_{p^2-1}) = \zeta_{p^2-1}^p$. This is not the same as $\zeta_{p^2-1}^{-1}$, except when $p=2$.
As a concrete example, take $p=3$. The $8$-th cyclotomic polynomial is $x^4+1$. It factors in $\mathbb{Q}_3$ as $(x^2+\sqrt{-2}x+1)(x^2-\sqrt{-2}x+1)$. So the Galois symmetry of $K$ permutes the two roots of $(x^2+\sqrt{-2}x+1)$, which are $\zeta_8$ and $\zeta_8^3$.