Let $L$ be a dihedral quartic field. That is, we have $[L : \mathbb{Q}] = 4$ and the Galois closure $M$ of $L$ is a degree 8 extension of $\mathbb{Q}$ with Galois group isomorphic to the dihedral group $D_4$ of order 8. It is well-known (see the diagram on page 6 of this paper: https://arxiv.org/abs/1704.01729) that $M$ contains five subfields which are degree 4 extensions of $\mathbb{Q}$; say $L = L_1, L_2, L_0, L_3, L_4$. In particular, $L_1, L_2$ are Galois conjugates and contain a common quadratic field $K_1$, $L_0$ is a Galois extension of $\mathbb{Q}$ and has Galois group isomorphic to $C_2 \times C_2$, while $(L_3, L_4)$ is related to $(L_1, L_2)$ via an *outer automorphism* which comes from the group theory of $D_4$, which we denote by $\phi$. In this language, we identify $L_3 = \phi(L_1), L_4 = \phi(L_2)$ and $L_3, L_4$ are Galois conjugates.

Given $L$, it is of course possible (say via the primitive element theorem) to find a quartic polynomial $f$ such that for some root $\alpha$ of $f$ we have $L = \mathbb{Q}(\alpha)$.

My question is, given a quartic polynomial $f$ such that for some root $\alpha$ of $f$ we have $L = \mathbb{Q}(\alpha)$ is a dihedral quartic field, how to find a polynomial $g$ whose coefficients are algebraic functions of the coefficients of $f$ such that for some root $\beta$ of $g$ we have $\phi(L) = \mathbb{Q}(\beta)$? Moreover, can one take advantage of the invariant theory of the homogenized binary quartic form $F(x,y) = y^4 f(x/y)$? That is, is $g$ naturally given by some quartic covariant of $F$?