Point reflection uniqueness Suppose we have normed vector space V and mapping R from V to itself satisfying the following properties:
1) R has unique fixed point $~a\in V$
2) $\forall x\in V ~~~~ |Rx-a|=|x-a|$
3) $\forall x \in V ~~~~ RRx = x$
Reflection across the point a obviously satisfy this properties but I wonder whether they uniquely define map R or there may be many functions satisfying them.
Thanks in advance.
 A: At least in Euclidean space $\Bbb R^n$, those properties don't quite uniquely identify a point-reflection. If the space has dimension at least $2$, we can pick any $2$-dimensional plane $W$ containing $a$ and the rotation of $\pi$ radians in $W$ about $a$ will satisfy those properties as well.
In $\Bbb R^2$, such a rotation is uniquely determined. But in $\Bbb R^n$ with $n \geq 3$, there are infinitely many such rotations.
A: Even in $\Bbb{R}^2$ you can easily construct a (non-linear) map $R$ with those properties, other than the point reflection map.  Let $R(a) = a$ and pick some arbitrary direction $d$.  For $x \neq a$, describe $x$ as $(r, \phi) : |\vec{x} -  \vec{a} | = r, \angle (d, \vec{x} -  \vec{a}) = \phi$.  (This is just saying let's describe $x$ in polar coordinates.  
Now choose some $ 0 < \alpha < \frac{1}{\pi}$ and define the map $R$ as $R(a) = a$ and:
$$
R( (r,\phi) = \left\{ \begin {array}{cl}
\left(r,  \phi + \alpha(\pi-\phi)^2 + \pi \right) & \text{if } 0 \leq \phi < \pi \\ \left(r,  \phi - \alpha\phi^2 + \pi \right) & \text{if } 0 \leq \phi < \pi
\end{array} \right. 
$$
It is easy to see that one side of direction $d$ maps into the other, so that $a$ is the only fixed point, and that $R^2(x) = x$. And since $R$ maps any circle centered onto $a$ to itself, it preserves distance from $a$.  Therefore it meets your conditions, but since $\alpha \neq 0$ it is not the point reflection map.
