This is how I see it. $G$ has 3 2-Sylow subgroups, and 4 3-Sylow subgroups. So $G$ acts on the 3-Sylow subgroups by conjugation, resulting in a map $\phi:G → S_4$. Let $K$ be the kernel of this map. Let $P$ be a 3-Sylow subgroup, and restrict $\phi$ to P; call the map $\phi_P$. Then the Sylow Permutation Theorem* states that the only fixed point of $\phi_P$ on the 4 conjugates is $P$ itself. This means that $\phi_P$ acts nontrivially on the other three 3-Sylow subgroups, implying that $K$ cannot contain an element of order 3, so that the order of $K$ is not divisible by 3, so it has to be 1, 2, 4, or 8.
Let $\psi: G → S_3$ be the action of $G$ on the 2-Sylow subgroups. Then $G$, order 24, acts on $S_3$ of order 6, implying a normal subgroup $V$ in $G$ of order 4. This subgroup then combines with a 3-Sylow subgroup, by semi-direct product, to form a subgroup $A$ of $G$ of order 12. There can be only one such subgroup of order 12, as there is no room for any more. So $A$ is of order 12 without normal 3-Sylow subgroups, so it is isomorphic to $A_4$ and $V$ is isomorphic to the Klein four-group. The 2-Sylow subgroups are of order 8, and each one of them must contain $V$ as a subgroup. Hence we have 1 identity element, 3 elements in $V$, 4 x 3 or 12 elements in the 2-Sylows outside of $V$, and 8 elements of order 3, accounting for all 24 elements of G. There is no more room for any more.
$K$ cannot be of order 8, because the subgroups of order 8 are not normal. $K$ cannot be of order 2, since in that case $K$ would be normal in $G$, so it combines with $P$ to form a subgroup of order 6 with subgroups of order 2 which are normal, hence is $Z_6$, containing elements of order 6, for which there is no room. Finally, $K$ can't be of order 4, since the normalizer of $P$ and its conjugates is of order 6, so the intersection of all these normalizers has to have order 1, 2, 3, or 6 and be contained in $K$, so it can't have order 4.
Therefore, $K$ is the trivial group, so that $\phi$ is an isomorphism between $G$ and $S_4$.
*The Sylow Permutation Theorem says that if a group $G$ has $n$ $p$-Sylow subgroups, then $G$ acts on $Syl_p(G)$ by conjugation, and the only fixed point of this action restricted to $P$ is $P$ itself. This occurs throughout the literature, mainly in proving that there are no simple groups of some specified order, but not as a theorem by itself.
Assume $f$ acts as identity on $P_i$ and as inversion on $P_j$. Let $d$ be the point fixed by $P_i$ and $c$ the point fixed by $P_j$, and $a,b$ the other two. Then
$$f((a\,c)(b\,d))=f(\underbrace{(a\,b\,c)}_{\in P_i}\underbrace{(a\,b\,d)}_{\in P_j})= (a\,b\,c)(a\,d\,b)= (a\,d\,c),$$
which is impossible (order 2 $\ne$ order 3).
And if $f$ acts as inversion on both $P_i$ and $P_j$,
$$ f((c\,b\,d))=f((a\,c\,b)(a\,b\,d))=(a\,b\,c)(a\,d\,b)=(a\,d\,c),$$
so $f$ maps $\langle (b\,c\,d)\rangle$ to $\langle (a\,c\,d)\rangle$, contradiction.
We conclude that the only possibility is that $f$ acts as identity on all $P_i$.
Best Answer
Assuming that you know that $f$ fixes every $3$-cycle, it follows that $f$ also fixes every even permutation (because $A_4$ is generated by $3$-cycles).
Now $A_n$ is the commutator subgroup of $S_n$ when $n \geq 3$ (proof: write a $3$-cycle as the commutator of two transpositions). So the parity of an permutation is invariant under an automorphism of $S_n$.
Thus $f$ permutes the transpositions $(12), (13), (14), (23), (24), (34)$.
Now $(123) = f(123) = f((13)(12)) = f(13)f(12)$ so $f(12)$ is $(12)$ or $(13)$ or $(23)$.
Also $(12)(34) = f((12)(34)) = f(12)f(34)$ so $f(12)$ is $(12)$ or $(34)$.
Thus $f(12)$ is $(12)$ hence $f = 1$.