The eight vertex permutations of a square form a non-abelian subgroup of $\operatorname{Sym}(\{1,2,3,4\})$ of order $8$, generated by a $4$-cycle and one diagonal swap

Powers of $r$, computed pointwise: $r^{2}$ sends $1 \mapsto 3$, $2 \mapsto 4$, $3 \mapsto 1$, $4 \mapsto 2$; $r^{3}$ sends $1 \mapsto 4$, $2 \mapsto 1$, $3 \mapsto 2$, $4 \mapsto 3$; and $r^{4}$ sends every point back to itself, so $r^{4} = \mathrm{id}$. None of $r$, $r^{2}$, $r^{3}$ is $\mathrm{id}$, each moving $1$. Hence $\operatorname{ord}(r) = 4$ and $r^{-1} = r^{3}$.

$s^{2} = \mathrm{id}$, since $s$ exchanges $1$ and $3$ and fixes $2$ and $4$, so applying it twice fixes every point; hence $s^{-1} = s$ and $s \ne \mathrm{id}$.

$sr = r^{3}s$. Both sides are computed pointwise: $s \circ r$ sends $1 \mapsto r(1) = 2 \mapsto s(2) = 2$, $2 \mapsto 3 \mapsto 1$, $3 \mapsto 4 \mapsto 4$, $4 \mapsto 1 \mapsto 3$; while $r^{3} \circ s$ sends $1 \mapsto s(1) = 3 \mapsto r^{3}(3) = 2$, $2 \mapsto 2 \mapsto 1$, $3 \mapsto 1 \mapsto 4$, $4 \mapsto 4 \mapsto 3$. The two agree at every point. With step 1.1 this gives $sr = r^{-1}s$, which with claim 1's other two equations completes claim 1.

$s \notin \{\mathrm{id}, r, r^{2}, r^{3}\}$: $s$ fixes $2$, while $r$, $r^{2}$ and $r^{3}$ send $2$ to $3$, $4$ and $1$ respectively, and $s \ne \mathrm{id}$ by step 1.2.

By induction from step 2.1, $s r^{k} = r^{-k} s$ for every $k \in \mathbb{Z}$: the case $k=0$ is trivial, and $s r^{k+1} = (s r^{k}) r = r^{-k} s r = r^{-k} r^{-1} s = r^{-(k+1)} s$, while applying $s$ on both sides of $s r^{k} = r^{-k}s$ and using $s^{2} = \mathrm{id}$ gives the statement for $-k$.

The eight listed elements are pairwise distinct. The four powers $r^{0}, r^{1}, r^{2}, r^{3}$ are pairwise distinct because $\operatorname{ord}(r) = 4$. If $r^{i}s = r^{j}s$ then $r^{i} = r^{j}$ by cancelling $s$ on the right, so $i = j$ for $i, j$ among $0,1,2,3$. And $r^{i}s = r^{j}$ would give $s = r^{j-i}$, contradicting step 2.2.

$D$ is closed under composition. A product of two listed elements has the form $(r^{i}s^{\varepsilon})(r^{j}s^{\delta})$ with $\varepsilon, \delta \in \{0,1\}$. If $\varepsilon = 0$ it equals $r^{i+j}s^{\delta}$; if $\varepsilon = 1$ then, moving $s$ past $r^{j}$ by step 3.1, it equals $r^{i}r^{-j}s\,s^{\delta} = r^{i-j}s^{1+\delta}$. In either case, reducing the exponent of $r$ using $r^{4} = \mathrm{id}$ and the exponent of $s$ using $s^{2} = \mathrm{id}$ gives one of the eight listed elements.

$D$ is closed under inverses: $(r^{i})^{-1} = r^{-i}$, again one of the four powers after reduction; and $(r^{i}s)^{-1} = s^{-1}(r^{i})^{-1} = s r^{-i} = r^{i}s$ by step 3.1 and $s^{-1} = s$, so each of the four elements $r^{i}s$ is its own inverse.

$D$ is not abelian: $sr = r^{3}s$ by step 2.1, while $rs = r^{1}s$; if these were equal then $r^{3} = r$ by cancelling $s$ on the right, contradicting the distinctness of the powers of $r$. This is claim 3.

$D$ is a subgroup: it contains $\mathrm{id}$, is closed under composition by step 4.1 and under inverses by step 4.2, so $x y^{-1} \in D$ for $x, y \in D$ and the one-step test applies. Its eight elements are distinct by step 3.2, so the map $8 \to D$ listing them is a bijection and $|D| = 8$.

$D = \langle \{r,s\} \rangle$: $D$ is a subgroup containing $r$ and $s$, so $\langle \{r,s\} \rangle \subseteq D$; conversely any subgroup containing $r$ and $s$ contains every $r^{i}s^{\varepsilon}$, hence contains $D$, so $D \subseteq \langle \{r,s\}\rangle$. This with step 5.1 is claim 2.

Every element of $D$ is a vertex symmetry. The map $r$ carries the four edges $\{1,2\}, \{2,3\}, \{3,4\}, \{4,1\}$ to $\{2,3\}, \{3,4\}, \{4,1\}, \{1,2\}$, so it maps $E$ onto $E$; being a bijection of $X$, it therefore also carries each of the two non-edges $\{1,3\}$, $\{2,4\}$ to a non-edge, and the "if and only if" holds. The map $s$ carries those four edges to $\{3,2\}, \{2,1\}, \{1,4\}, \{4,3\}$, again onto $E$, so the same applies. The vertex symmetries form a subgroup, since the defining condition is preserved by composition and, being an equivalence, by inverses; hence it contains $\langle \{r,s\}\rangle = D$.

Conversely let $f$ be a vertex symmetry. The neighbours of a point $x$, meaning the $y$ with $\{x,y\} \in E$, are exactly $r(x)$ and $r^{3}(x)$, and these two are distinct because $r^{2}(x) \ne x$ by step 1.1. There is a unique $i \in \{0,1,2,3\}$ with $r^{i}(1) = f(1)$, since those four values are $1, 2, 3, 4$; put $f_1 = r^{-i}f$, again a vertex symmetry, with $f_1(1) = 1$.

Since $\{1,2\} \in E$, the pair $\{f_1(1), f_1(2)\} = \{1, f_1(2)\}$ is an edge, so $f_1(2)$ is a neighbour of $1$, that is $f_1(2) \in \{2,4\}$. Since $\{2,3\} \in E$, the point $f_1(3)$ is a neighbour of $f_1(2)$, and it differs from $f_1(1) = 1$ because $f_1$ is injective; the neighbours of $2$ are $1$ and $3$ and the neighbours of $4$ are $3$ and $1$, so in both cases $f_1(3) = 3$. Then $f_1(4)$ is the one element of $X$ not already taken.

So either $f_1$ is $1 \mapsto 1$, $2 \mapsto 2$, $3 \mapsto 3$, $4 \mapsto 4$, that is $f_1 = \mathrm{id}$ and $f = r^{i}$; or $f_1$ is $1 \mapsto 1$, $2 \mapsto 4$, $3 \mapsto 3$, $4 \mapsto 2$. In the second case $f_1 = r^{2}s$, since $r^{2} \circ s$ sends $1 \mapsto s(1) = 3 \mapsto r^{2}(3) = 1$, $2 \mapsto 2 \mapsto 4$, $3 \mapsto 1 \mapsto 3$ and $4 \mapsto 4 \mapsto 2$; hence $f = r^{i}r^{2}s = r^{i+2}s \in D$.

By steps 7.1, 9.1 and 10.1 the vertex symmetries of the square are exactly the elements of $D$, which is claim 4; claims 1, 2 and 3 are steps 2.1, 6.1 and 4.3.