Determinant conjecture for twisted derivations when the conductor is $2^r p$

Let $\mathbb{K}=\mathbb{Q}(\zeta)$ be the $n$th cyclotomic number field with $n=2^r p$, where $r\in\mathbb{N}$ and $p$ is an odd rational prime, and let $O_{\mathbb{K}}=\mathbb{Z}[\zeta]$. Let $u,v\in U(n)$ with $u\ne v$, and let $\sigma,\tau:O_{\mathbb{K}}\to O_{\mathbb{K}}$ satisfy $\sigma(\zeta)=\zeta^u$ and $\tau(\zeta)=\zeta^v$. Write

$$\beta=\sum_{i=0}^{2^{r-1}(p-1)-1}b_i\zeta^i\in O_{\mathbb{K}},\qquad \beta(\tau-\sigma)(\zeta)=\sum_{i=0}^{2^{r-1}(p-1)-1}\left(\sum_{j=0}^{2^{r-1}(p-1)-1}a_{ij}b_j\right)\zeta^i,$$

and let $A=[a_{ij}]$. Suppose $v-u=2^{e_1}p^{e_2}m$, where $m$ is a positive integer, $e_1,e_2$ are non-negative integers, and $2,p$ do not divide $m$. Determinant conjecture. The matrix $A$ is a $2^{r-1}(p-1)\times 2^{r-1}(p-1)$ matrix with determinant $\pm2^{2^{e_1}(p-1)}$ if $1\le e_1\le r-1$ and $e_2\ge1$, determinant $\pm p^{2^{r-1}}$ if $e_1\ge r$ and $e_2=0$, and determinant $\pm1$ otherwise. The claim is intended to characterize the coefficient matrix governing inner $(\sigma,\tau)$-derivations in this cyclotomic setting; the sign depends on the order in which the rows are formed, and the conjectural determinant underlies the subsequent criteria for inner and outer derivations.