It is a well known result from Galois theory that, given a Galois extension $L$ of a field $K$, an element $x$ of $L$ is in $K$ if for all $\sigma$ in $Gal(L/K)$, one has $\sigma(x)=x$. 


My question is: is it true that every element $y$ of $L$, such that there exists at least one field homomorphism that doesn't preserve $y$, is in $K$ if and only if for all element $s$ of $Aut(L)$, $s(y)=y$ implies $s\in Gal(L/K)$?  

Thanks in advance.



About some kind of "converse" of a theorem from Galois theory
No. Take $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$, $L=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, $y=\sqrt{3}$, and $s$ the automorphism of $L$ that fixes $\sqrt{3}$ and $\sqrt{5}$, but sends $\sqrt{2}$ to $-\sqrt{2}$.


