If we look at the cross ratio $(x_0 x_1:x_2 x_3) = \lambda$ of 4 points in projective space, I can see that by looking at all possible permutations (24 of them) of the points we can see that only 6 of them give the same cross ratio.
I.e: $\lambda, \frac{1}{\lambda}, 1-\lambda, \frac{1}{1-\lambda}, \frac{\lambda}{\lambda -1}, \frac{\lambda-1}{\lambda}$
Is there a more efficient way to see this rather than calculating all 24 permutations?
Write the cross product as
$$ \lambda = \frac{(x_1-x_3)(x_2-x_4)}{(x_2-x_3)(x_1-x_4)} $$
As we have the permutation group $S_4$ we only need the generators $(12)$, $(13)$ and $(14)$.
Now
$$(12) \lambda = \frac{(x_2-x_3)(x_1-x_4)}{(x_1-x_3)(x_2-x_4)} = \frac{1}{\lambda}$$ $$(13) \lambda = \frac{(x_3-x_1)(x_2-x_4)}{(x_2-x_1)(x_3-x_4)} = \frac{\lambda}{\lambda - 1} $$ $$(14) \lambda = \frac{(x_4-x_3)(x_2-x_1)}{(x_2-x_3)(x_4-x_1)} = 1 - \lambda $$
What we find is that
$$1 \lambda = 1$$ $$(12)(34) \lambda = \lambda$$ $$(13)(24) \lambda = \lambda$$ $$(14)(23) \lambda = \lambda$$
which is the Klein-Four group $V_4$,
So we can write $S_n = V_4 \otimes S_3$
$V_4$ has 4 elements and $S_3$ has 6 elements. The $S_3$ gives the
$$\lambda, \frac{1}{\lambda}, 1 - \lambda, \frac{1}{1-\lambda}, \frac{\lambda}{1-\lambda}, \frac{1-\lambda}{\lambda}$$
Total we have $24 = 4 \times 6$.
A conceptually organized way to see all the calculation is the following:
The cross ratio of $x_1,x_2,x_3,x_4$ is the unique $\lambda$ such that the 4-point configuration $(x_1,x_2,x_3,x_4)$ is equivalent to $(\infty,0,1,\lambda)$ by Möbius transform. (There are other conventions, but it always involve moving $x_1,x_2,x_3$ to three standard points and call the resulting fourth point the cross ratio.)
From this viewpoint, the effect of permuting $(x_1,x_2,x_3)$ is easy to figure out by writing down the unique Möbius transform that takes $(\infty,0,1)$ to a permutation of it. This pins down all permutations of $S_4$ that fixes 4. So you will have a table with 6 entries containing things like $(x_2x_1:x_3x_4)=1/\lambda$, etc.
Then we use the property that if we exchange two pairs of points simultaneously, the cross ratio does not change. A statement and a conceptual proof is as follows.
Theorem. The configuration $(x_1,x_2,x_3,x_4)$ is equivalent to $(x_2,x_1,x_4,x_3)$ under Möbius transform.
Proof. We claim that any Möbius transform that swaps two points must be an involution. If the claim is proved, consider the unique Möbius transform $T$ that exchanges $x_1,x_2$ and take $x_3$ to $x_4$. Since it is an involution, $T(x_4)=T(T(x_3))=id(x_3)=x_3$, so we are done.
To prove the claim, let $S$ be any Möbius transform that swaps $x_1,x_2$. Work over complex numbers, $S$ must have a fixed point (either by the existence of solution to a quadratic equation, or by the existence of an eigenvector for any $2\times 2$ matrix defining $S$), call it $x_0$, which must be different from $x_1,x_2$. Now since $S$ swaps $x_1,x_2$ and fixes $x_0$, $S^2$ must fix $x_0,x_1,x_2$. But any Möbius transform that fixes three distinct points must be the identity map, so $S^2=id$ as claimed. QED
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Note that the proof clearly works for other two switching $(13)(24)$ and $(14)(23)$ as well.
With these two in mind, one can easily convert every permutation to one of the 6 entries in your table. For example, say we want $(x_3x_4:x_2x_1)$. Then we shall permute $x_4$ to the last place. Recall the simultaneously swapping rule and the table entry, we get $$(x_3x_4:x_2,x_1)=(x_2x_1:x_3x_4)=1/\lambda.$$
