Let $K = \mathbb{Q}_3$. Let $L=K(\alpha)$ given by $\min_K(\alpha) = x^4-3x^2+18$. This extension is cyclic of degree $4$ with ramification index $2$. In particular, the intertial degree is also $2$. See also this page.
Let $L'/L$ be the unique unramified extension of degree $2$. Since $L'/L$ is generated by a primitive $5$-th root of unity $\zeta_5$, it is $L'= K(\alpha,\zeta_5)$.
If I am not wrong, the Galois group of $L'/K$ is $$ \operatorname{Gal}(L'/K) \simeq C_4 \times C_2 = \langle \sigma' \rangle \times \langle \sigma \rangle. $$
Suppose that $\sigma: \alpha \mapsto -\alpha, \, \zeta_5 \mapsto \zeta_5$ (which has order $2$) and $\sigma': \alpha \mapsto \alpha, \, \zeta_5 \mapsto \zeta_5^3$ (which has order $4$).
According to the Fundamental Theorem of Galois Theory, it is $Gal(L/K) = Gal(L'/K)/Gal(L'/L)$. A generator of $Gal(L'/L)$ is given by $\alpha \mapsto \alpha, \, \zeta_5 \mapsto \zeta_5^9 = \zeta_5^{-1}$.
If I look at the explicit elements of the quotient, they must be $\{ Id, \bar{\sigma}, \bar{\sigma'}, \bar{\sigma} \bar{\sigma'} \}$ (it is $(\sigma')^2 \in Gal(L'/L)$) But none of these elements seem to be a generator of the quotient. (if my computations are correct, all these elements have order $2$ at most). Since $L/K$ is indeed cyclic, this cannot be possible.
I assume I made a mistake with identifying my elements. Could you please say where my line of thought is wrong?
You can show that $$L = \Bbb{Q}_3( \zeta_8, \zeta_{16} 3^{1/2}) \qquad \implies \qquad L' = \Bbb{Q}_3( \zeta_{80}, 3^{1/2})$$ Then it is clear that $$Gal(L'/\Bbb{Q}_3)=Gal(L'/\Bbb{Q}_3(3^{1/2}))\times Gal(L'/\Bbb{Q}_3(\zeta_{80}))= C_4\times C_2$$ and $Gal(L'/L) = \langle (2,1)\rangle$.
Any element of $Gal(L'/\Bbb{Q}_3)$ whose reduction is the Frobenius will have order $4$ both in $Gal(L'/\Bbb{Q}_3)$ and $Gal(L/\Bbb{Q}_3)$