$[F:K]$ finite , $Aut_K(F)=\{f_1,...,f_n\}$ , there is $a\in F$ such that $\{f_1(a),...,f_n(a)\}$ forms a basis for $F$ over $K$ ; is $F/K$ Galois? Let $F/K$ be a finite extension field , let $Aut_K(F)=\{f_1,...,f_n\}$ , suppose there exist $a\in F$ such that $\{f_1(a),...,f_n(a)\}$ forms a basis for  $F$ over $K$ ; then is $F/K$ Galois extension ? 
 A: Yes. The given conditions imply that $ [F : K] \leq |\textrm{Aut}(F/K)| $, and this is a necessary and sufficient condition for a finite extension of fields to be Galois (in fact, this is how being Galois is defined in some sources).
A: Let $G = Aut(F/K), m = |G|, n = [F:K]$ and $L$ be the subfield of $F$ fixed by $G$. Then $F/L/K$ is a tower of extension and $[F:L] = r$ and $[L:K] = n/r$. Also, writing $F$ as a $L$-algebra, we see that any automorphism of $F/L$ extends to an automorphism of $F/K$, therefore $Aut(F/L) \subset Aut(F/K)$ and by the definition of $L$ : $Aut(F/L)= G$.
For any $c \in F$, let $f_c(x) = \prod_{\sigma \in G} (x-\sigma(c))$ which is a polynomial of $F[x]$. Note that each of its coefficients are fixed by $G$, therefore $f_c \in L[x]$. Thus, the minimal polynomial of $c$ over $L$ divides $f_c$ whose roots are $\sigma(c)\in F$. 
This means that $F/L$ is Galois and $r = [F:L] = |Aut(F/L)| = m$ (write $L = F(b_1,\ldots,b_k)$ and use induction on $k$)

Assume $m=n$, then $L=K$ and $F/K$ is Galois.

If $\{\sigma(a)\}_{\sigma \in G}$ is a $K$-vector space basis of $F$, then $n=[F:K] \le m$ and by the preceding $n=m$ which means that $F/K$ is Galois
