Follow-up to question: Aut(G) for G = Klein 4-group is isomorphic to $S_3$

Every group is a set (together with a binary operation on the set). Every automorphism of a group $G$ is a bijective function from the underlying set of $G$ to itself (which in addition respects the operation of $G$). So every automorphism of a group $G$ can be viewed as a subgroup of the group of all permutations on the underlying set of $G$; in fact, since an automorphism must send the identity of $G$ to itself, you can even view every automorphism of $G$ as a permutation of the set $G-\{e\}$.

Here you are getting a bit confused because you are viewing your group $G$ as a subgroup of $S_4$ (nothing wrong with that), and then you are trying to understand $\mathrm{Aut}(G)$ (which can be viewed as a subgroup of $S_{G-{e}} \cong S_3$) as acting on the set $\{1,2,3,4\}$ as well. While some automorphisms can be defined in terms of an action on $\{1,2,3,4\}$, not every automorphism can.

If you view $G$ as the set $\{\mathrm{id}, (1\;2), (3\;4), (1\;2)(3\;4)\}$, then letting $$\begin{align*} x &= (1\;2),\\ y &= (3\;4),\\ z &= (1\;2)(3\;4)\\ e &= \mathrm{id} \end{align*}$$ then the automorphisms of $G$ will always map $e$ to itself, and so you can view the automorphisms as being elements of $S_{\{x,y,z\}}$, the permutation group of $\{x,y,z\}$. The elements of the automorphism group are then (written in 2-line format): $$\begin{align*} \mathrm{id}_{\{x,y,z\}} &= \left(\begin{array}{ccc} x & y & z\\ x & y & z \end{array}\right)\\ f_1 &= \left(\begin{array}{ccc} x & y & z\\ y & z & x \end{array}\right)\\ f_2 &=\left(\begin{array}{ccc} x & y & z\\ z & x & y \end{array}\right)\\ f_3 &= \left(\begin{array}{ccc} x & y & z\\ y & x & z \end{array}\right)\\ f_4 &= \left(\begin{array}{ccc} x & y & z\\ z & y & x \end{array}\right)\\ f_5 &= \left(\begin{array}{ccc} x & y & z\\ x & z & y \end{array}\right) \end{align*}$$ and you can verify that each of them is an automorphism; since every automorphism corresponds to a unique element of $S_{\{x,y,z\}}$, and every element of this group is an automorphism, then this is the automorphism group.

However, you are trying to view your group $G$ as a subgroup of $S_4$. Can the automorphisms of $S_4$ be "induced" by some permutation of $\{1,2,3,4\}$? Equivalently:

If we view the Klein $4$-group $G$ as the subgroup of $S_4$ generated by $(1\;2)$ and $(3\;4)$, is every automorphism of $G$ induced by conjugation by an element of $S_4$?

Well, it doesn't. The reason it doesn't is that $x$, $y$, and $z$ don't all have the same cycle structure: conjugation by elements of $S_4$ (or more precisely, by elements of the normalizer of $G$ in $S_4$) will necessarily map $z$ to itself, because $z$ is the only element with its cycle structure (product of two disjoint transpositions) in $G$. So the only automorphisms that can be viewed as coming from "acting on ${1,2,3,4}$" are the identity and $f_3$

Added: However: it is possible to view the Klein $4$-group as a different subgroup of $S_4$: identify $x\mapsto (1\;2)(3\;4)$; $y\mapsto (1\;3)(2\;4)$, $z\mapsto (1\;4)(2\;3)$. It is not hard to verify that this is a subgroup of order $4$, and since it has three elements of order $2$, it is isomorphic to the Klein $4$-group. Now you can indeed realize every automorphism of $G$ via conjugation by an element of $S_4$ (in several different ways). Here's one way: the automorphism $\mathrm{id}$ can be realized as conjugation by the identity. The automorphism $f_1$ is given by conjugation by $(2\;3\;4)$; $f_2$ is given by conjugation by $(2\;4\;3)$; the automorphism $f_3$ is given by conjugation by $(2\;3)$; $f_4$ is given by conjugation by $(2\;4)$; and $f_5$ is given by conjugation by $(3\;4)$. You can now easily see that the automorphism group is indeed isomorphic to $S_3$, since it corresponds precisely to the subgroup of $S_4$ that fixes $1$ (you can take any of the other four one-point stabilizers as well).