No I don't think this is right. For example the extension $\Bbb{Q}(\sqrt{3},\sqrt{2})$ is a Galois extension of $\Bbb{Q}$ because it is the splitting field of $(x^2 - 3)(x^2-2)$. This polynomial has exactly 4 distinct roots in $\Bbb{Q}(\sqrt{3},\sqrt{2})$ and it can easily be shown that its Galois group is the Klein 4-group $V_4$. But $V_4$ has order 4 that is not equal to $4!=24$.
Now you are thinking that the Galois group has to be $S_4$. Let me tell you why this is not possible. Let us call $\sqrt{3}$ root #1, $\sqrt{2}$ root #2, $-\sqrt{3}$ root #3, $-\sqrt{2}$ root #4. Given a cycle in $S_4$ let that cycle act on the roots simply by permuting the numbers $1,2,3$ and $4$. For example, the cycle $(12)$ exchanges $\sqrt{3}$ and $\sqrt{2}$ and keeps the negative guys fixed. But then this cannot possibly be a valid element of the Galois group because:
Elements of the Galois group must send for example $\sqrt{2}$ to another root of the minimal polynomial of $\sqrt{2}$ over $\Bbb{Q}$. The minimal polynomial of $\sqrt{2}$ when viewed as an element of $\Bbb{Q}(\sqrt{2},\sqrt{3})$ is $x^2 -2$. Therefore the only possibility for where $\sqrt{2}$ can be sent to is $-\sqrt{2}$, because this is the only other root of this polynomial in any splitting field. So therefore the cycle $(12)$ above cannot be a valid element of the Galois group (if we view the Galois group as sitting inside of $S_4$).
It follows that the Galois group in this case can only be viewed as a proper subgroup of $S_4$ and hence cannot have order $4!=24$.
Edit: Since you seem to be having some trouble understanding the Galois group, let me explain a bit more here. Now I assume that you know what an $F$ - algebra is (otherwise how would you understand field extensions?)
The following is the start of how one describes the Galois group:
Let $A = F[x]$ where $F$ is a field. Let $\iota_A$ denote inclusion of $F$ into $A$. Then any $F$ - algebra homomorphism from $F[x]$ to some other $F$ - algebra $B$ (where the homomorphism in question for $B$ is just the inclusion map $\iota_B : F \to B$) is completely determined by specifying the image of $x$ in $B$. This is because if we have an $F$ - algebra homomorphism $\varphi$ from $F[x] \rightarrow B$, we must have that
$$\iota_B = \varphi \circ \iota_A.$$
In particular this means that $\varphi$ must be the identity on the coefficients of a polynomial in $F[x]$. This explains why $\varphi$ is completely determined by its action on $x$. Now we claim that we have a bijection of sets
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x],B\big)\Big\} \longleftrightarrow B $$
where the bijection is given by $f$ that maps $\varphi$ on the left to $\varphi(x)$ with inverse $g$ that maps an element in $b \in B$ to the homomorphism $\varphi_b$ which is evaluation at $b$. Viz. $\varphi_b$ is just the homomorphism that sends $x$ to $b$. You can check that $f$ and $g$ are mutual inverses.
Now a corollary of this is that we have a bijection
$$\Big\{\operatorname{Hom}_{\text{$F$ -algebra}} \big(F[x]/(f(x)),B\big)\Big\} \longleftrightarrow \Big\{b\in B : \varphi_b(f(x)) = 0 \Big\}. $$
I will leave you to work out the details of how this comes from the fact I stated before. Essentially it is due to the universal property of quotient rings that says given a unique ring homomorphism $\varphi$ from $F[x]$ to $B$ we have unique ring homomorphism from the quotient of $F[x]/(\ker \varphi)$to $B$ such that $\varphi$ factorises through the quotient. I can edit my post to explain this more if you wish.
This is the start of how one gets a description of the Galois group because giving a homomorphism from some field say $F(\alpha)$ to itself (which is automatically an automorphism by the Rank - Nullity Theorem) is by my description above equivalent equivalent to specifying a root of the minimal polynomial of $\alpha$ over $F$ in $B$. But then our $B$ here is exactly what we started with, that is $F(\alpha)$ so that $\alpha$ must be sent to another root of its minimal polynomial over $F$.
Does this help to explain more to you?
Begin by seeing that
$$\mathbb{Q}(-3\sqrt{6}+3\sqrt[3]{5})=\mathbb{Q}(\sqrt{6},\sqrt[3]{5}).$$
Now, this is not a normal field, since the minimal polynomial for $\sqrt[3]{5}$ does not split. The normal closure of this would then need to be
$$\mathbb{Q}(\sqrt{6},\sqrt[3]{5},\omega)$$
where $\omega$ is a cube root of unity.
This is an extension of degree $12$ over $\mathbb{Q}$, and so we have $12$ automorphisms that fix $\mathbb{Q}$. We can see that these automorphisms are just the combinations of automorphisms moving around the roots of $X^3-1$, $X^3-5$, and $X^2-6$.
In particular, $\sqrt{6}$ can be mapped to $\sqrt{6}$ or to $-\sqrt{6}$.
$\sqrt[3]{5}$ can be mapped to $\omega\sqrt[3]{5}$, $\omega^2 \sqrt[3]{5}$ or $\sqrt[3]{5}$.
Since the Galois Group acts transitively on the roots of the minimal polynomial, this gives us all of the possibilities.
Best Answer
As explained in my comment, to conclude that all $4$ admissible permutations of the roots give rise to automorphisms, it suffices to show that the degree of the splitting field $K$ of $(x^2-2)(x^2-3)$ over $\mathbb{Q}$ has degree $4$. Clearly $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. You have a tower $\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2})\subseteq K$. The first extension is degree $2$ because $X^2-2$ is irreducible by Eisenstein. Note that $K$ is obtained from $L:=\mathbb{Q}(\sqrt{2})$ by adjoining a root of $X^2-3$. Thus the degree of $L$ over $K$ is at most $2$. If it is equal to $1$, then $X^2-3$ splits over $L$. Thus you can write $\sqrt{3}=a+b\sqrt{2}$ for rational numbers $a$ and $b$. Squaring gives $3=a^2+2ab\sqrt{2}+2b^2$. Since $\{1,\sqrt{2}\}$ is a $\mathbb{Q}$-basis for $L$, this forces $a^2+2b^2=3$ and $2ab=0$, so $a=0$ or $b=0$. If $a=0$, then $2b^2=3$ so $3/2$ is a rational square, which is false. Similarly, $b=0$ leads to $a^2=3$ with $a$ rational, another contradiction. Thus $X^2-3$ can't split in $L$, so it's irreducible over $L$, and $K/L$ has degree $2$, giving $[K:\mathbb{Q}]=2\cdot 2=4$.
If for some reason you don't want to use the fact that $3/2$ is not a rational square, you could use the (non-trivial) fact that $a$ and $b$ must actually be integers, and then $a^2+2b^2=3$ forces $a$ and $b$ to each be $\pm 1$, while $2ab=0$ forces one of them to be zero, again a contradiction.