I think we agree that two (squarefree) quadratic extensions of $\mathbb Q$, say $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ are not isomorphic.
Now consider the following tower of fields
$L=\mathbb Q(i,\sqrt[4]3,\sqrt[3]3,\sqrt[3]2)$
$M_1= \ K(\sqrt[3]3), M_2 = \ K(\sqrt[3]2), M_3 = K(\sqrt[3]6), M_4 = K(\sqrt[3]{12})$
$ K= \mathbb Q(i,\sqrt[4] 3)$
I think there are no more intermediate fields.
We know that $\mathrm{Gal}(K/\mathbb Q)= D_4$ (sometimes also denoted $D_8$, the dihedral group of the square), and $\mathrm{Gal}(L/K)= \mathbb Z/3\times\mathbb Z/3$.
In my notes it says:
'All $M_i$ are clearly isomorphic with isomorphism given by $\sigma$, the $4$-cyclic element of $D_4$.'
I don't understand why this is the case.
You are right $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ are not isomorphic. If $\varphi$ was an isomorphism between $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ you would have $\varphi(\sqrt 2) = a + b \sqrt 3$ with $a,b \in \mathbb Q$ and therefore $(a + b \sqrt 3)^2 - 2=0$ which cannot happen as $\sqrt 2$ and $\sqrt 3$ are irrational.
The $M_i$ are not isomorphic for similar arguments. Are you sure that the statement is not about Galois groups of the $M_i$?