Let $\Gamma = \mathbb{Q}(\sqrt[5]n)$ a pur quintic field and $k= \mathbb{Q}(\sqrt[5]n,\zeta_5)$ its normal closure. Let $\operatorname{Gal}(k/\Gamma)= \langle\tau\rangle$ with $\tau^4 = 1$ and $\operatorname{Gal}(k/\mathbb{Q}(\zeta_5))= \langle\sigma\rangle$ with $\sigma^5 = 1$. we have the relations : $\sigma\tau = \tau\sigma^2$ and $\tau\sigma = \sigma^3\tau$. let $\mathcal{A}$ a class verify : $\mathcal{A}^\tau = \mathcal{A}$. Using the fact that $\sigma^4+\sigma^3+\sigma^2+\sigma^1=0$, we prove that the class $\mathcal{A}^{\sigma^3+2\sigma^2+3\sigma-1}$ is stable under the action of $\langle\sigma\rangle$, wich we call ambiguous class.
My question is, how using all this relations we can proof that $(\mathcal{A}^{\sigma^3+2\sigma^2+3\sigma-1})^\tau = \mathcal{A}^{1-3\sigma-2\sigma^2-\sigma^3}$
You probably made a typo somewhere.
$\tau\sigma\tau^{-1}=\sigma^3$ and $A=A^\tau$ and $\sigma^4+\sigma^3+\sigma^2+\sigma^1+1=0$ gives
$$(A^{\sigma^3+2\sigma^2+3\sigma-1})^\tau= A^{\tau (\sigma^3+2\sigma^2+3\sigma-1)}=A^{\tau(\sigma^3+2\sigma^2+3\sigma-1)\tau^{-1}}=A^{\sigma^9+2\sigma^6+3\sigma^3-1}=A^{2\sigma^3 - \sigma^2 + \sigma - 2}$$