Let $|K:F| = p$ be a field extension of degree $p$. For each $Aut(K/F)\ni\sigma \neq id$, denote $K^\sigma$ the the fixed field of $\sigma$ and we have $$p = |K:F| = |K:K^\sigma||K^\sigma:F|$$ because $p$ is prime and $\sigma \neq id$, we see that $K^\sigma = F$, so indeed $F$ is the fixed field for each non identity element in $Aut(K/F)$, and we have $K$ over $F$ is Galois (this last part is a standard result, but non trivial).
Is this correct? It seems too simple to be true for me... and I didn't even have any assumptions for $Char(F)$.
There are discussions about this here, it seems to be complicated.