Denote the composite of all such fields by $E$ and consider the action of any $\tau\in Aut(L/F )$ on it. We see that any such $\tau$ sends $E$ into itself: Let $B_i=\left\{b_{i,1},b_{i,2},...b_{i,m}\right\}$ denote any set of elements that span $\sigma_iK$ over $F$. Define $P=\left\{p_1,p_2,...p_s\right\}$ to be the set of all possible products of elements $b_{i,r}$ in the union $B_1 \cup B_2 \cup ...B_q$. The set $P$ spans $E$ over $F$. So, for arbitrary $\alpha \in E$ we have $$\alpha=\sum_{j=0}^sa_jp_j $$ $a_j \in F$ applying some $\tau \in Aut(L/F)$ and noting that $\tau$ fixes the $a_j$'s, we get $$\tau\left(\alpha\right)=\sum_{j=0}^sa_j\tau\left(p_j\right)$$
Each $p_j$ is a product of $b$'s in the union above and each $b_{i,r}$ is in some $\sigma_{i}K$ so $\tau$ sends it to an element of $\tau\sigma_{i}K$ which is an element of $E$ by definition. So, the image of a product of any number of $b$'s will still be in $E$ since all the images are in $E$. So, the image of $p_j$ is some $\beta$ again in $E$ which may or may not be in $P$, but this doesn't matter since $\tau\left(\alpha\right)$ is in $E$ anyway and we started with an arbitrary $\alpha$. Noting that $\tau$ can't send two elements to the same element since it's an automorphism of the larger field $L$, I conclude that any such $\tau$ is an automorphism of $E$ fixing $F$. But the number of embeddings $N$ of $E$ into an algebraic closure which contains $L$ that fix $F$ is the order of the set of cosets $\left|Aut\left(L/F\right)\right|/\left|Aut\left(L/E\right)\right|$ by the fundamental theorem. Combining these results we get, $$\frac{\left|Aut\left(L/F\right)\right|}{\left|Aut\left(L/E\right)\right|}\ge\left|Aut\left(E/F\right)\right|\ge\left|Aut\left(L/F\right)\right|$$ which (since $L/E$ is galois) leads to $$\left|Aut\left(L/E\right)\right|=\left[L:E\right]=1$$
Is this correct?