I presume you mean the conjugation action of $G=S_4$ on the set of its subgroups. In that case, the stabilizer of $H$ is $N_G(H)$, the normalizer of $H$.
You are correct that every element of $H$ stabilizes $H$. In fact, since $H$ has order $4$, it is abelian, so conjugation by an element $h\in H$ not only normalizes $H$, it centralizes $H$ (every element of $H$ is unchanged by conjugation by $h$).
But there are additional elements of $S_4$ which normalize $H$. I claim that we can find an element $g \in S_4$ such that conjugation by $g$ exchanges $(12)$ and $(34)$ while fixing $id$ and $(12)(34)$. Then $g$ normalizes $H$ but does not centralize $H$, so $g \not\in H$.
What element $g$ can we use to exchange $(12)$ and $(34)$? For example, $g=(13)(24)$ will accomplish this: check that
$$g(12)g^{-1} = (34)$$
and
$$g(34)g^{-1} = (12)$$
Of course $g(id)g^{-1} = id$, and finally,
$$g(12)(34)g^{-1} = g(12)g^{-1}g(34)g^{-1} = (34)(12) = (12)(34)$$
So now we have at least five elements of $G$ which normalize $H$, namely the four elements in $H$, plus the element $g$ defined above. Since the normalizer is a subgroup of $G$, its order must divide $|S_4| = 24$. Since $5$ does not divide $24$, that means the normalizer contains additional elements that we have not found yet. See if you can find the rest.
Another way to see that the stabilizer (normalizer) of $H$ is more than just $H$ is to look at the orbit of $H$ under conjugation. See if you can recognize that these three subgroups are all conjugate to each other:
$$\{id, (12), (34), (12)(34)\} = H$$
$$\{id, (13), (24), (13)(24)\}$$
$$\{id, (14), (23), (14)(23)\}$$
and since conjugation preserves cycle structure, these are the
only conjugates of $H$. This means that the orbit of $H$ has size $3$. By the orbit-stabilizer theorem, the size of the orbit of $H$ is the same as the index of its stabilizer (normalizer). Therefore, $|G : N_G(H)| = 3$, which means that $|N_G(H)| = 8$. Above, we identified $5$ elements of $N_G(H)$, so this argument shows that there are $3$ more.
Best Answer
$(1\,4)$ is a representative of the left coset… $(1\,4)H$.
Also, two permutations $\rho,\sigma\in S_4$ define the same coset modulo $H$ if and only if $\rho^{-1}\sigma\in H$, i.e. if the composition $\rho^{-1}\sigma$ leaves $4$ fixed. Now you can easily check this condition is not satisfied by any of the pairs $\{\operatorname{Id}, (1\,4), (2\,4), (3\,4)\}$, e.g. $$(1\,4)(2\,4)=(1\,4\,2)$$ Finally, observe $H$ is isomorphic to $S_3$ (it permutes $1,2,3$ between them). So the set $S_4/H$ has cardinal $$\bigl|S_4/H\bigr|=\frac{\bigl|S_4\bigr|}{\bigl|S_3\bigr|}=4,$$ so we've obtained the exact number of cosets representatives.