Isometry fixing two points of a geodesic line Let $H$ be a hyperbolic space, and let $\Gamma \subset H$ be a geodesic line, i.e., the image of an isometry from $\mathbb{R}$ to $H$. If $f$ is an isometry of $H$ that fixes two distinct points of $\Gamma$, is it true that $f(\Gamma) = \Gamma$?
 A: The answer is yes. This is a simple application of the fact isometries send geodesics to geodesics, and every geodesic can be uniquely defined by two points in the union of the hyperbolic plane and its boundary circle.
If you've not seen either of these facts justified, feel free to ask and I'll try to spend the time explaining them.

To help clarify why every two points define a unique geodesic, I'll explain for the special case when both points $z$ and $w$ are in the hyperbolic plane $\mathbb{H}$. First, recall that geodesics in the hyperbolic upper half plane are either vertical Euclidean straight lines or semicircles with center on the real axis. Recall also that for every pair of points in $\mathbb{H}$, there exists at least one geodesic which passes through both points.
Suppose $z$ and $w$ share the same real coordinate and differ on the imaginary coordinate. Then, there are no semicircles with center on the real line which pass through both $z$ and $w$ and so it must be a vertical straight line which passes through them. It should be clear that such a line is unique.
Now, suppose that $z$ and $w$ differ on their real coordinate. There are a couple of ways that we could show that there is a unique geodesic which passes through both $z$ and $w$. One way would be to show that there is a unique $p\in\mathbb{R}\subset\partial\mathbb{H}$ such that $|z-p|=|w-p|$. This can be shown algebraically. You can also see this geometrically by drawing small circles around $z$ and $w$ of the same radius and then letting them grow continuously. It should be clear that there will be exactly one time when these two circles intersect on the real axis and they intersect at exactly one point.
The reason we want $|z-p|=|w-p|$ is because the point $p$ satisfies the necessary condition for the center of the semi circle passing through both $z$ and $w$. As $p$ is unique, and the length $|z-p|$ has to be the radius of such a semicircle, the geodesic passing through both $z$ and $w$ is uniquely determined by these two numbers.
The other way to see that a unique geodesic passes through $z$ and $w$ is to notice that there exists an isometry of the hyperbolic plane which maps both $z$ and $w$ to the imaginary axis. To describe this isometry, first take any geodesic $\Gamma_0$ which passes through both points (we know at least one geodesic passes through them) and call the end points $a$ and $b$ with $a<b$. Let $f$ be the isometry given by $$f(x)=\frac{x-b}{x-a}.$$ The isometry $f$ maps $b$ to $0$ and $a$ to $\infty$ and so must map $\Gamma_0$ to $\{x\:|\:\mbox{Re}(x)=0\}$ the imaginary axis. As $z,w\in\Gamma_0$, we also know that $f(z)$ and $f(w)$ are on the imaginary axis. By the previous discussion, $f(\Gamma_0)$ is the unique geodesic which passes through $f(z)$ and $f(w)$ but then $f^{-1}(f(\Gamma_0))$ must be the unique geodesic passing through $z$ and $w$ and so $\Gamma_0$ is unique.
From a quick search, you can find a proof of this as well in these lecture notes.
