Let $k= \mathbb{Q}(\sqrt[5]n, \zeta_5)$ be the normal closure of $\Gamma = \mathbb{Q}(\sqrt[5]n)$ and Kummer's extension of the cyclotomic field $\mathbb{Q}(\zeta_5)$. Let $Gal(k/\mathbb{Q})=\langle\sigma, \tau\rangle $, such that $Gal(k/\mathbb{Q}(\zeta_5))=\langle\sigma\rangle $ and $Gal(k/\Gamma)=\langle\tau\rangle $. Its known that $\Gamma$ has four conjugates denoted $\Gamma', \Gamma'', \Gamma''', \Gamma''''$.
Let $k^{(1)}, \Gamma^{(1)}, (\Gamma')^{(1)}, (\Gamma'')^{(1)}, (\Gamma''')^{(1)}$ and $ (\Gamma'''')^{(1)}$ be respectively the Hilbert class field of $k, \Gamma, \Gamma', \Gamma'', \Gamma'', \Gamma'''$and $\Gamma''''$. We Suppose that $Gal(k^{(1)}/k) \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} $ then the extension $k^{(1)}/k$ has 6 intermediates fileds denoted $K_i$ $1\leq i \leq 6$.
I proved that $k\Gamma^{(1)}, k(\Gamma')^{(1)}, k(\Gamma'')^{(1)}, k(\Gamma''')^{(1)}, k(\Gamma'''')^{(1)}$ and the relative genus filed $(k/\mathbb{Q}(\zeta_5))^*$ are the six fields $K_i$ when the radicand $n$ is not divisible by any prime $p\equiv 1 (mod 5)$
we order the subfields $K_i$ as follows:
$K_1=k\Gamma^{(1)}, K_2 = k(\Gamma')^{(1)}, K_3 = k(\Gamma'')^{(1)}, K_4=k(\Gamma''')^{(1)}, K_5=k(\Gamma'''')^{(1)}$ and $K_6=(k/\mathbb{Q}(\zeta_5))^*$
I need to know what is the relation between the fildes $K_1, K_2, K_3, K_4, K_5$ under the action of $Gal(k/\mathbb{Q})=\langle\sigma, \tau\rangle $.