Okay, I'll answer your question on the cross ratio:
First of all, recall its definition
$\lambda = \lambda(z) = [z,z_2,z_3,z_4] = \frac{z-z_3}{z - z_4} : \frac{z_2 - z_3}{z_2 - z_4}$
which is the unique Möbius transformation $z \mapsto \lambda(z)$ sending the ordered triple $(z_2,z_3,z_4)$ to $(1,0,\infty)$—it is a general and easy fact that a Möbius transformation is determined by the images of three distinct points.
In view of the symmetries you display in your question, the only thing we have to figure out are the Möbius transformations that arise from permuting the three entries $(z_2,z_3,z_4)$ in the cross ratio. For this we need only post-compose our initial cross ratio $\lambda$ with the unique Möbius transformation sending $(1,0,\infty)$ to some permutation of $(1,0,\infty)$.
For instance, the cross-ratio $[z,z_4,z_3,z_2]$ is obtained from $\lambda$ by postcomposing it with the Möbius transformation sending the triple $(1,0,\infty)$ to $(\infty,0,1)$. This is easy: Since $1$ is sent to $\infty$, we must have a transformation of the form $\lambda \mapsto \frac{a\lambda + b}{\lambda - 1}$ (with $a \neq -b$ but we won't need that additional information). As $0$ is sent to $0$, we must have $b = 0$ and since $\infty$ is sent to $1$ we must have $a = 1$, and our sought transformation is
$[z,z_4,z_3,z_2] = \frac{\lambda}{\lambda - 1}.$
Similar considerations lead to the entire list of symmetries given on Wikipedia.
In order to save you some work, I'll give one more indication: swapping the last two entries corresponds to $(1,0,\infty) \mapsto (1,\infty,0)$, so simply $\lambda \mapsto \frac{1}{\lambda}$. Hence, there really only remains one more thing to do: figure out the transformation sending $(1,0,\infty)$ to $(\infty,1,0)$ and this is just as easy as the case I just did. You'll get the expression for $[z,z_3,z_4,z_2]$ and after that hardly any further computation is required.