So I was reading in Abstract Algebra about the extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$. It was proven that the degree of this extension is four, so there are at most four elements in Gal($\mathbb{Q}(\sqrt{2},\sqrt{3})$/$\mathbb{Q}$). The automorphism were defined by sending $\sqrt{2}$ to $\pm\sqrt{2}$ and $\sqrt{3}$ to $\pm\sqrt{3}$.
Lets look at the map that sends $\sqrt{2}$ to $\sqrt{3}$ and $\sqrt{3}$ to $\sqrt{2}$. If we look at an element a + b$\sqrt{2}$ + c$\sqrt{3}$ + d$\sqrt{6}$ ant apply this map we get a + b$\sqrt{3}$ + c$\sqrt{2}$ + d$\sqrt{6}$. To me this looks like an automorphism, why isn't it?
It isn't an automorphism of $\Bbb Q(\sqrt2, \sqrt 3)/\Bbb Q$, and isn't even a field automorphism at all, because it can't respect multiplication. Call your map $\varphi$. Consider the fact that $\sqrt2 \cdot \sqrt2 = 2$. If this were an automorphism, we would have $$ \varphi(2) = \varphi(\sqrt2\cdot \sqrt2)=\varphi(\sqrt2)\cdot \varphi(\sqrt2) = \sqrt3\cdot\sqrt3 = 3\neq 2 $$ But for any field automorphism, we need to have $2= \varphi(2)$. So no field automorphism could send $\sqrt2$ to $\sqrt 3$, much less an automorphism of $\Bbb Q(\sqrt2, \sqrt 3)/\Bbb Q$.
Your map is certainly an automorphism on the vector space $\Bbb Q(\sqrt2, \sqrt 3)$ over the rationals (which is a big hint that any potential flaw must involve the field multiplication in $\Bbb Q(\sqrt2, \sqrt 3)$). But that's not enough to be a field automorphism.