Let $p$ be a prime, $m \in \mathbb{Z}^+$ such that $m \equiv 3 \pmod 4$ and $p \equiv 1 \pmod m$, $\zeta_m$ be a primitive $m$-th root of unity, $K = \mathbb{Q}(\zeta_{2m}) = \mathbb{Q}(\zeta_m)$, and $\mathcal{P}$ a prime of $K$ above $p$. If $G = $ Gal$(K/\mathbb{Q})$ then $\sigma_j$ acts by $\zeta_m \mapsto \zeta_m^j$ (where $\gcd(j,m) = 1$). Then we write $\mathcal{P}_j = \sigma_j(\mathcal{P})$.
We have a multiplicative character $\psi : \mathcal{O}_{K}/ \mathcal{P} \to \mathbb{C}^{\times}$ of order $2m$ modulo $\mathcal{P}$. We denote \begin{align*} J(\psi) = \sum_{a \in \mathbb{F}_{p^f}} \psi(a) \psi(1-a) \end{align*} to be the Jacobi sum for $\psi$. Denote $0 \leq L(j) < 2m$ to be reduction of $j$ modulo $2m$, and for $1 \leq i < m$ we define \begin{align*} S_i(m) = \{ j \,:\, 0< j < 2m; \,\gcd(j,2m) = 1;\, L(ji) < m \}. \end{align*} Then, from Theorem 2.1.14 in Berndt, Evans and Williams' Gauss and Jacobi Sums book, the prime factorization of $J(\psi^i)$ is \begin{align*} J(\psi^i)\mathcal{O}_{K} = \prod_{j \in S_i(m)} \mathcal{P}_{j^{-1}}. \end{align*}
Here is what I'm not getting: If $\sigma_k \in G$, then $\sigma_k(J(\psi^i)) = J(\psi^{ki})$ and so by the Theorem, \begin{align*} J(\psi^{ki})\mathcal{O}_{K} = \prod_{j \in S_i(m)} \sigma_k(\mathcal{P}_{j^{-1}}) = \prod_{j \in S_i(m)} \mathcal{P}_{kj^{-1}}. \end{align*}
However, looking at an explicit example with $K = \mathbb{Q}(\zeta_{14})$, I get that $J(\psi)\mathcal{O}_K = \mathcal{P}_1\mathcal{P}_3\mathcal{P}_5$ and $J(\psi^2)\mathcal{O}_K = \mathcal{P}_1\mathcal{P}_5\mathcal{P}_{4}$. But $\sigma_2(\mathcal{P}_1\mathcal{P}_3\mathcal{P}_{5}) = \mathcal{P}_2\mathcal{P}_6\mathcal{P}_{3}$. I can't figure out what is going wrong here!
I think I see now what the issue is. Since I am looking at the multiplicative character $\psi$ of order $2m$, then $\psi(a)$ is a $2m$-th roof of unity for some $a \in \mathcal{O}_K/\mathcal{P} = \mathbb{F}_p$. If $\sigma_{j}$ is a map such that $\sigma_{j}(\zeta_{2m}) = \zeta_{2m}^j$, we have $\sigma_{2k}(\zeta_{2m}) = \sigma_{2k}(-\zeta_m) = \zeta_m^{2k}$. But at the same time, $\sigma_{m + 2k}(\zeta_{2m}) = -\zeta_m^{2k}$.
The problem was that when I define the automorphisms in $G$ by their action on $\zeta_{2m}$, $\sigma_{2k} \notin G$, but instead $\sigma_{m + 2k} \in G$ (which acts as $\sigma_{2k}$ when the automorphisms are defined by what they do to $\zeta_7$).
So, with $m = 7$, I sill have $J(\psi)\mathcal{O}_K = \mathcal{P}_1\mathcal{P}_3\mathcal{P}_5$, but this time (with $9$ instead of $2$) $J(\psi^9)\mathcal{O}_K = \mathcal{P}_3\mathcal{P}_9\mathcal{P}_{13}$, and indeed, $\sigma_9(J(\psi)) = \mathcal{P}_3\mathcal{P}_9\mathcal{P}_{13}$.