For completeness I give some definitions. Let $p$ a prime number and consider $V$ a 2-dimensional $\mathbb{F}_p$-vector space. Consider $k$ a sub-algebra of $\mathrm{End}(V)$ that is a field of $p^2$ elements. Then $C=k^* \subset \mathrm{GL}_2(V)\simeq\mathrm{GL}_2(\mathbb{F}_p)$ is called a non-split Cartan subgroup.
Let $\mathcal{N}$ be the normalizer of $C$ in $\mathrm{GL}_2(\mathbb{F}_p)$. Then each $s \in \mathcal{N}$ induce an automorphism of $k$ defined as $$ x \mapsto sxs^{-1}$$ In particular the kernel of the map $\mathcal{N} \longrightarrow Aut(k)$ is $C$. Now, $k$ has only one non-trivial automoprhism that is the Frobenius automorphism $\sigma: a \mapsto a^p$, $a \in k$.
So, can I say that each $s \in \mathcal{N} \setminus C$ has the form $\sigma x$ with $x \in C$?
The idea come from the fact that $\sigma$ is an invertible $\mathbb{F}_p$-linear map of $\mathbb{F}_{p^2} \simeq k$, hence representable as a matrix of $\mathrm{GL}_2(\mathbb{F}_p)$ when we chose a basis for $\mathbb{F}_{p^2}$ over $\mathbb{F}_p$.
Ok, I think that this work. But it can be write better.
The Frobenius automorphism of $\mathbb{F}_{p^2}$ is an invertible $\mathbb{F}_p$-linear map. If we fix a basis of $\mathbb{F}_{p^2}$ over $\mathbb{F}_p$, then $\sigma$ is representable as a matrix of $\mathrm{GL}_2(\mathbb{F}_p)$. We denote with $S_F$ the associated matrix of $\sigma$. Fix an $s \in \mathcal{N}$, as we have seen it induces an automorphism $\varphi_s$ on $k$, in particular it is induced by the Frobenius automorphism of $\mathbb{F}_{p^2}$. So $\varphi_s$ acts on $C$ as conjugation by $S_F$. Then we have $sxs^{-1}=S_F^{-1}x S_F$ for all $x\in C$. Thus the element $S_F s$ commute with all elements of $C$. However $C$ coincides with its centralizer in $\mathrm{GL}_2(\mathbb{F}_p)$ and since $s \notin C$ then it is in the non-trivial coset of $C$ in the subgroup $<C,S_F>$.