As the name suggests, we are equivalently finding isometries $F:\mathbb{R}^2\rightarrow \mathbb{R}^2$ with respect with the Manhattan metric (also called $L^1$ I think) where we define $d(A,B)=||\vec{AB}||_1=||\vec{B}-\vec{A}||_1=|x_B-x_A|+|y_B-y_A|$, where $A=(x_A,y_A),B=(x_B,y_B)$ belong to $\mathbb{R}^2$. It is easy to see that if $F$ is an isometry, then we can compose it with a translation, or a central symmetry or a symmetry with respect to a line with slope $0,\pm1,\infty$. (In other words by doing something like $x\mapsto x+c, x\mapsto -x, x\mapsto y$). Therefore, assume wlog that $O$, the origin, maps to itself. Consider the square $A=(r,0),B=(-r,0),C=(0,r),D=(0,-r)$. We have $d(O,A)=d(O,B)=1=d(O,C)=d(O,D)$ and $d(A,B)=2r=d(C,D)$. Therefore, $F(A),F(B),F(C),F(D)$ still belong to the circle of radius $r>0$ centered at the origin (i.e. $C(0,r)=\{A\in\mathbb{R}^2: ||A||=r\}$). Since $A,B$ and $C,D$ are the only pair of points with distance $2r$, it follows that the pairs $\{A,B\}$ and $\{C,D\}$ get permuted. By the symmetries stated above, we may wlog assume that $A,B,C,D$ remain in place. Now if we pick a point $E=(x,r-x)$ with $x\in(0,r)$, we have $E\in C(0,r)$ and $d(A,E)=2x$ and $d(C,E)=2(r-x)$. It is therefore easy to see that $F(E)=E$(because there are no other points in $C(O,r)$ with these distances from $A$ and $C$). In other words, the circle stays fixed, up to $90$ degree rotations/reflection wrt the axes. This is true for all circles $C(O,r)$ with positive radius, but they may have a different symmetry with respect to their original circle. So, assume wlog the points $A_1,B_1,C_1,D_1$ remain fixed (these points are the same points defined as above for $r=1$) and consider the points $A_r,B_r,C_r,D_r$ (once again defined as above, but for arbitrary $r\in(0,\infty)-\{1\}$). We have $d(A_1,A_r)=|r-1|$. If $F(A_r)=B_r$, we'd have $d(F(A_1),F(A_r))=r+1>|r-1|$; if $F(A_r)=C_r$ (or $D_r$) we have once again $d(F(A_1),F(A_r))=r+1>|r-1|$. Therefore, $F(A_r)=A_r$ for all $r>0$, or in other words all circles centered at the origin have the same symmetry from them to their images wrt $F$. So $F=(f,g)$ can be in one of the following families a)$(f,g)=(\varepsilon_1x+a,\varepsilon_2y+b)$ b)$(f,g)=(\varepsilon_1y+a,\varepsilon_2x+b)$ Where $a,b\in\mathbb{R}$ and $\varepsilon_1,\varepsilon_2\in\{-1,1\}$.

https://hackmd.io/@sine/TMO_22P3 I am lazy to translate it to English