Consider the field $E = \mathbb{Q}(\sqrt{3},i,\sqrt{2})$; $E$ is an extension of $\mathbb{Q}$
$(1)$ Find a basis of $E$ over $\mathbb{Q}$
$(2)$ For each of the automorphisms $\sigma$ below, find all extensions of $\sigma$ to an isomorphism mapping $E$ to a subfield of $\overline{\mathbb{Q}}$
$\circ$ $\sigma :$ $\mathbb{Q}(\sqrt{3},i)\rightarrow\mathbb{Q}(\sqrt{3},i)$ where $\sigma(i)=i$ and $\sigma\sqrt{3}=-\sqrt{3}$
$\circ$ $\sigma :$ $\mathbb{Q}(\sqrt{2})\rightarrow\mathbb{Q}(\sqrt{2})$ where $\sigma(\sqrt{2})=\sqrt{2}$
$\circ$ $\sigma :$ $\mathbb{Q}(\sqrt{6}i)\rightarrow\mathbb{Q}(\sqrt{6}i)$ where $\sigma(\sqrt{6}i)=-\sqrt{6}i$.
For part (1) I found a basis: $\{ 1, i, \sqrt{3}, \sqrt{2},\sqrt{6}, \sqrt{2}i, \sqrt{3}i, \sqrt{6}i\}$
For part (2) I am attempting the first map.
$\psi(i)=i, \psi(\sqrt{3})=-\sqrt{3}$
For the second map:
$\psi(\sqrt{2}) =\sqrt{2}$ ie: $\psi={1\!\!1}$
And for the third map...
$\psi_{1}(\sqrt{6})=-\sqrt{6},\psi_{1}(i)=i$
$\psi_{2}(\sqrt{6})=\sqrt{6}, \psi_{2}(i)=-i$
Am I doing this correctly? Any hints/assistance would be greatly appreciated.