Let $K/k$ be a Galois extension, and let $F$ be an intermediate field between $k$ and $K$. Let $H$ be the subgroup of $\text{Gal}(K/k)$ mapping $F$ into itself. Show that $H$ is the normalizer of $\text{Gal}(K/F)$ in $\text{Gal}(K/k)$.
Remark: This problem has appeared in MSE couple of times but all these problems have a bit different condition. More precisely, they consider "mapping $F$ onto itself", i.e. for $\sigma \in H$ it follows that $\sigma(F)=F$.
My solution: Denote this normalizer by $N$. We have to show that $H=N$.
I was able to show that $N\subseteq H$ without any difficulties.
Let show the converse, namely $H\subseteq N$.
Take $\sigma \in H$ then $\sigma \in \text{Gal}(K/k)$, $\sigma(F)\subseteq F$. We have to show that that $\sigma \in N$, i.e. $$\sigma \text{Gal}(K/F)\sigma^{-1}=\text{Gal}(K/F).$$ So basically speaking we have to show double containment.
i) Take $\tau\in \text{Gal}(K/F)$ then we see that $\sigma^{-1}\tau \sigma\in \text{Gal}(K/k)$. For any $x\in F$ we see that $\sigma^{-1}\tau \sigma(x)=\sigma^{-1}(\sigma(x))=x$ here I've used that $\sigma(x)\in F$ and $\tau$ fixes $F$ pointwise. So $\sigma^{-1}\tau \sigma\in \text{Gal}(K/F)$ $\Rightarrow$ $\tau \in \sigma \text{Gal}(K/F)\sigma^{-1}$. Therefore, $\text{Gal}(K/F)\subseteq\sigma \text{Gal}(K/F)\sigma^{-1}$.
ii) Let $\tau \in \sigma \text{Gal}(K/F)\sigma^{-1}$ $\Rightarrow$ $\tau=\sigma\hat{\tau}\sigma^{-1},$ where $\hat{\tau}\in \text{Gal}(K/F)$.
Easy to see that $\sigma\hat{\tau}\sigma^{-1}\in \text{Gal}(K/k)$.
Take any $x\in F$ then $\tau(x)=\sigma\hat{\tau}\sigma^{-1}(x)$.
If $\sigma(F)=F$ then I can easily show this case.
But we have that $\sigma(F)\subseteq F$ and in this case we have some troubles because in this case $\sigma^{-1}(x)$ may not be in $F$.
I would be very grateful if anyone can show how to proceed this!