The eight vertex permutations of a square form a non-abelian subgroup of $\mathrm{Sym}(\{1,2,3,4\})$ 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 $r^4=\mathrm{i}\mathrm{d}$. None of $r$, $r^2$, $r^3$ is $\mathrm{i}\mathrm{d}$, each moving $1$. Hence $\mathrm{ord}(r)=4$ and $r^{-1}=r^3$.

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

$sr=r^{3}s$. 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.

$s\mathrm{\notin }\{\mathrm{i}\mathrm{d},r,r^2,r^3\}$: $s$ fixes $2$, while $r$, $r^2$ and $r^3$ send $2$ to $3$, $4$ and $1$ respectively, and $s\mathrm{\ne }\mathrm{i}\mathrm{d}$ 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 $s{r}^{k+1}=(s{r}^k)r=r^{-k}sr=r^{-k}{r}^{-1}s=r^{-(k+1)}s$, while applying $s$ on both sides of $s{r}^k=r^{-k}s$ and using $s^2=\mathrm{i}\mathrm{d}$ 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 $\mathrm{ord}(r)=4$. 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 $(r^i{s}^{\epsilon })(r^j{s}^{\delta })$ with $\epsilon ,\delta \in \{0,1\}$. If $\epsilon =0$ it equals $r^{i+j}{s}^{\delta }$; if $\epsilon =1$ 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 $r^4=\mathrm{i}\mathrm{d}$ and the exponent of $s$ using $s^2=\mathrm{i}\mathrm{d}$ 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 $\mathrm{i}\mathrm{d}$, 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 $\mathrm{\mid }D\mathrm{\mid }=8$.

$D=⟨\{r,s\}⟩$: $D$ is a subgroup containing $r$ and $s$, so $⟨\{r,s\}⟩\subseteq D$; conversely any subgroup containing $r$ and $s$ contains every $r^i{s}^{\epsilon }$, hence contains $D$, so $D\subseteq ⟨\{r,s\}⟩$. 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 $⟨\{r,s\}⟩=D$.

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 $r^2(x)\mathrm{\ne }x$ 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 $f_1=\mathrm{i}\mathrm{d}$ 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.