Properties of an unique involution for $2$ points in $\mathbb{R}P^1$ Consider a projective line $\mathbb{R}P^1$. An involution $\phi$ is defined as a projective transformation such that $\phi^2 = I_d \neq \phi$.
I want to proof that there exist for $2$ points $A,B \in \mathbb{R}P^1$ an unique involution if it has $A$ and $B$ as fixed point.
For another property consider that unique involution. It should be valid that for a point $P \in \mathbb{R}P^1$, different from $A$ and $B$, the pairs $\{A,B\}$ and $\{P, \phi(P)\}$ should seperate each other harmonically. So $(A,B,P,\phi(P)) = -1$. How can one show this?
For both properties, I do not really have a clue how it can be proven.
 A: I'll prove the two statements in reverse order using the second to prove the first.
1. A point $P$ and its image $\phi(P)$ are harmonic conjugates with respect to the two fixed points $A$ and $B$ of an involution, i.e the cross ratio $(A,B; P,\phi(P)) = −1$.
We know that in an involution if $Q$ is the image of $P$ then $P$ is the image of $Q$ under the same involution. We also know that the cross ratio of four collinear points and the cross ratio of their images under any projectivity are equal. As an involution is a projectivity, we get.
$$(A, B; P, Q) = (A, B; Q, P).$$
Now, if in a set of four collinear points, the second pair is interchanged, the cross ratio of the resulting arrangement is the reciprocal of the cross ratio of the original arrangement. So, 
$$(A, B; P, Q) = \frac{1}{(A, B; Q, P)}.$$
For both of these statements to simulatneously hold, we must have
$$(A, B; P, Q)^2 = 1.$$
This implies that $(A, B; P, Q)$ must be either +1 or -1. It can't be 1 because that implies $P$ and $Q$ are the same point which is contrary to our assumption. Hence the cross ratio $(A, B; P, Q)$ or $(A, B; P, \phi(P))$ must be -1.
2. An involution is uniquely defined by two fixed points
We know that any projectivity between two lines is uniquely determined when three distinct points on the second line are assigned as the respective images of three arbitrary, distinct points on the first line.
In this case the two lines are the same and we have been given two points and their images - the fixed points. We only need one more point and its image on the same line to uniquely define the involution. We can use the result we just proved above to get this pair.
Take any point $P$ which is not one of the fixed points of the involution. We know from the previous result that $P$ and its image $\phi(P)$ must form a harmonic conjugate with respect to the fixed points $A$ and $B$ of the involution. So, we can easily find the image of $P$ using the cross ratio equation
$$(A, B; P, \phi(P)) = -1.$$
Now we have three pairs of distinct points and their images which is enough to determine a unique projectivity. Hence the two fixed points of the involution are enough to uniquely determine the involution.
