If $a$ , $b$ have the same minimal polynomial there exists a field automorphism such that $\sigma(a)=b$ Two elements are said to be conjugates if they have the same minimal polynomial.
It is true that for every extension $L/K$, $a\in L$ and $b$ conjugate to $a$ in $L$ there exists an element $\sigma \in \text{Gal}(L/K)$ such that $\sigma(a)=b$?
I know that this is true if $a$ is a primitive element and I guess that is true if the extension $L/K$ is normal (because you only need to find a $\sigma \in \text{Gal}(K(a,b)/K)$ such that $\sigma(a)=b$ and as the extension is normal you can extend it to a $\tilde{\sigma} \in \text{Gal}(L/K)$), but in the general case I don't even have an intuition of what happens.
Can anyone give a counterexample or show a proof in the normal/general case?

This is false.  For instance, consider $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt[4]{2})$, $a=\sqrt{2}$, and $b=-\sqrt{2}$.  Then $a$ and $b$ are conjugate over $K$, but no automorphism of $L$ can send $a$ to $b$ since $a$ has a square root in $K$ but $b$ does not.
It is true if $L$ is normal over $K$.  In that case, there is an isomorphism $\sigma:K(a)\to K(b)$ sending $a$ to $b$, which extends to an isomorphism $\tilde{\sigma}:L\to L$ by normality of $L$.
