$L/K$ is a normal field extension, $H=\operatorname{Aut}(L/K)$ prove that $L/L^{H}$ is separable extension and so $ L/L^{H}$ is a Galois extension.
first $L^{H} =\bigl\{a \in L \mid \forall \sigma \in H , \sigma(a)=a \bigr\}$
I want to look in $\operatorname{Aut}(L/L^H) = [\tau \mid \tau (L^H)=L^H] $ so for all those previous statements I know that $\tau =\sigma$. but somehow I need to show that for every irreducible polynomial $f(x) \in L^H [x]$ , there is only one way to map every root of $f(x)$ in $L$ and show there are $n=|\operatorname{Aut}(L/L^H)|$ roots, so $f(x)$ will be separable over $L$.
Take $\alpha \in L$ and consider the set $\{\sigma(\alpha) \mid \sigma \in H\}$ and let the distinct elements be $\{\alpha_1, \dots \alpha_n \}$. Then consider:
$$h(x) = \prod_{i=1}^n (x-\alpha_i)$$
Then it's not hard to see that $h$ is fixed by $H$, as it only permutes the factors on the right side. Hence we have that $h(x)$ has coefficients in the fixed field of $H$. In particular $h(x) \in L^H[x]$. Also we can prove that it's the minimal polynomial of $\alpha$ over $L^H$. Indeed if $g = \min(\alpha,L^H)$, then we have that $\sigma(\alpha)$ is a root too and so it has factors $(x-\alpha_i)$ in $L$ and so $h \mid g$ and by the irreducibility $h=g$. Moreover by the way we have constructed $h$ is separable and so is $\alpha$ over $L^H$. Hence $L^H \subseteq L$ is a separable extension.
For the normality we can prove that $\text{Gal}(L/K) = \text{Gal}(L/L^H)$ and in particular it's a normal subgroup of $\text{Gal}(L/K)$ and so we have that $L^H \subseteq L$ is normal.