For (fintie) Galois extension $K/F$, it is easy to show that $N_{K/F}\left(\frac{\beta}{\sigma\left(\beta\right)}\right)=1$ for all $0\ne\beta\in K$, $\sigma\in\text{Gal}\left(K/F\right)$. I want to show it's conditional reverse, with assumpstion $\text{Gal}\left(K/F\right)=\left<\sigma\right>$.
In other words, I want to show "If $\text{Gal}\left(K/F\right)=\left<\sigma\right>$, $N_{K/F}\left(\alpha\right)=1$ for given $0\ne\alpha\in K$, then $\alpha=\frac{\beta}{\sigma\left(\beta\right)}$."
You are searching for a non-zero fixed point of the $F$-linear map $$\phi: K \to K,\beta \mapsto \alpha\sigma(\beta).$$
The assumption $N(\alpha)=1$ makes sure that we have $\phi^n=\operatorname{Id_K}$ (where $n$ is the degree of $K/F$). In particular the minimal polynomial of $\phi$ in $F[T]$ is a divisor of $$T^n-1=(T-1)(1+ T + \dotsb + T^{n-1}).$$
By the well known fact that $1,\sigma, \dotsc, \sigma^{n-1}$ are linear independent, we obtain $1+ \phi + \dotsb + \phi^{n-1} \neq 0$, hence the minimal polynomial contains the factor $T-1$, which is equivalent to the statement that $1$ is an eigenvalue of $\phi$, the desired result.