Can an isometry of the hyperbolic plane that maps a circle to a disjoint circle have a fixed point? Can an isometry of the hyperbolic plane that maps a circle (centred on the real line) to a disjoint circle (also centred on the real line) have a fixed point? By disjoint, I mean that the two circles have no points in common. If we were considering Moebius transformations with complex coefficients then of course this would be true. But I can't decide the truth when the coefficients are real.

An example is the matrix $\pmatrix{0 &1\\\\-1 & 1}$ acting by the fractional linear transformation $f(z) = 1 / (-z+1)$. The restriction of $f$ to the upper half plane is an isometry, it fixes the point $p = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ in the upper half plane, and the two points $p$ and its complex conjugate $\bar p$ are the only two points in $\mathbb{C} \cup \\{\infty\\}$ that are fixed by $f$.

In fact for any fractional linear transformation $f$ and any point $q \in \mathbb{C}$ such that $q \
e f(q) \in \mathbb{C}$, by continuity there exists $\epsilon>0$, and a circle $C$ centered on $q$ and contained in the Euclidean $\epsilon$-ball $B(q,\epsilon)$ whose image $f(C)$ is a circle contained in the Euclidean $\epsilon$-ball $B(f(q),\epsilon)$, where $\epsilon$ is chosen so tiny that $B(q,\epsilon) \cap B(f(q),\epsilon) = \emptyset$.