A linear symmetry

Linearity. $s(\lambda(x,y))=(\lambda y,\lambda x)=\lambda s(x,y)$ and $s\big((x,y)+(x',y')\big)=(y+y',\ x+x')=s(x,y)+s(x',y')$.

Involution. $s(s(x,y))=s(y,x)=(x,y)$, so $s\circ s=\mathrm{id}$.

Eigenspaces. $s(v)=v\iff y=x$: $E_1=\operatorname{span}\big((1,1)\big)$. $s(v)=-v\iff y=-x$: $E_{-1}=\operatorname{span}\big((1,-1)\big)$. The vectors $(1,1)$ and $(1,-1)$ are independent, so $\mathbb{R}^2=E_1\oplus E_{-1}$ and $s$ is diagonalizable with eigenvalues $1$ and $-1$.

Conclusion. $s$ is a linear involution (the reflection across the line $y=x$, parallel to the direction $(1,-1)$).