Let $F$ be a field and let $\overline{F}$ an algebraic closure of $F$. Let $x , y \in \overline{F}$ and let $x' , y' \in \overline{F}$ be two roots of the irreducible, and monic, polynomials $p(x , F)(X) , p(y , F)(Y) \in F[X]$ ($p(x , F)(x) = 0 = p(y , F)(y)$). We also suppose that $$ [F(x , y) : F] = [F(x) : F] [F(y) : F] \tag{1} $$ and I have to show that there exist an $F$-automorphism $\sigma : \overline{F} \to \overline{F}$ such that $\sigma(x) = x'$ and $\sigma(y) = y'$.
Whithout using $(1)$, we can show that there are two $F$-automorphism ${\sigma}_1 , {\sigma}_2 : \overline{F} \to \overline{F}$ such that ${\sigma}_1(x) = x'$ and ${\sigma}_2(y) = y'$ and it is not too difficult: since $x' , y' \in \overline{F}$ are roots of $p(x , F)(X)$ and $p(y , F)(Y)$, respectively, there are two $F$-immersions ${\sigma}_1 , {\sigma}_2 : \overline{F} \to \overline{F}$ such that ${\sigma}_1(x) = x'$ and ${\sigma}_2(y) = y'$ ($i : F \hookrightarrow \overline{F}$ is obviously a $F$-immersion and then if $\sigma : \overline{F} \to \overline{F}$ is an extension of $i$, it is automatically a $F$-immersion) and then they are also $F$-automorphisms (all the $F$-immersions $K \to K$, where $K / F$ is an algebraic extension of fields, are $F$-automorphisms).
Using $(1)$, can we show that ${\sigma}_1 = {\sigma}_2$? I do not know but we can get the next statements: using the formule of the multiplicity for degrees and $(1)$, $$ [F(x) : F] [F(y) : F] = [F(x , y) : F] = [F(x , y) : F(y)] [F(y) : F]\mbox{,} $$ so we can remove $[F(y) : F]$ on both sides to see that $$ [F(x , y) : F(y)] = [F(x) : F] = \deg p(x , F)(X) $$ and analogously we have that $$ [F(x , y) : F(x)] = [F(y) : F] = \deg p(y , F)(X)\mbox{.} $$ Using that the amount of roots of $p(x , F)(X)$ must be the same of $F$-automorphism $\overline{F} \to \overline{F}$ (and the same for $p(y , F)(X)$), it should be enough to show that $\deg p(x , F)(X) = 1$ (or $\deg p(y , F)(X) = 1$). Can I deduce this? Thank you very much in advance.