Cross-ratio and projective transformations Let $P=\{p_1,\ldots,p_6\}\subset\mathbb{P}^1$ be a set of six general points of the projective line. In particular there are no two different subsets $\{p_{i_1},\ldots,p_{i_4}\}$ and $\{p_{j_1},\ldots,p_{j_4}\}$ of four elements of $P$ such that the $p_{i_k}$'s and the $p_{j_k}$'s have the same cross-ratio.
Assume that there exists a projective transformation $f:\mathbb{P}^1\rightarrow\mathbb{P}^1$ such that $f(P) = P$; that is $f$ acts on $P$ by permuting the $p_i$'s. 
Do we have then that $f = \operatorname{Id}_{\mathbb{P}^1}$ ?
 A: As Igor points out, the cross-ratio of a subset of four distinct points is not well-defined. If you meant to stipulate that all tuples of four distinct points from $P$ should have different cross-ratios, except for the unavoidable repetitions, then the answer is yes, you can conclude that $f$ is the identity transformation:
Since $f$ is a projective linear transformation, it preserves cross-ratios. In particular, the cross-ratio of $(f(p_1), f(p_2), f(p_3), f(p_4))$ is the same as that of $(p_1, p_2, p_3, p_4)$. Thus, by the hypothesis, $f$ in fact must permute already the set consisting of the first four points, in a manner which leaves the cross-ratio unchanged. If $f$ isn't the identity, then it must permute them in two pairs. Without loss of generality (renumbering points if necessary and conjugating everything with a suitable projective linear transformation to pin down the first three points) the only possibility is swapping $p_1=\infty$ with $p_2=0$ and $p_3=1$ with some $p_4$; thus $f\colon x\mapsto p_4/x$ and (edit if $f$ moves $p_5$) $p_6=p_4/p_5$; but then the cross-ratio of $(\infty,0,1,p_5)$ would equal that of its $f$-image $(0,\infty,p_4,p_6)$, a tuple supported on a different subset of $P$.
Edit - Or else, this $f$ fixes both $p_5$ and $p_6$, which must then be two distinct square roots of $p_4$ in the field we're working with. But then the cross-ratio of $(\infty,0,1,p_5)$ would equal that of its $f$-image $(0,\infty,p_4,p_5)$; again violating the hypothesis.
A: I think the answer depends on your definition of general position. Consider four points, whose cross ratio is $y.$ Then permutations will have such diverse cross ratios as $ \frac{1}{y}, 1-y, 1-\frac{1}{y}, \frac{1}{1-y}, \frac{y}{y-1}$. Setting one of these quantities equal to $y$ will give you a finite order projective transformation reducing to a permutation on the four points. It is not too easy to conjure up two other points which are permuted by this transformation.
