Are the only transformations which preserve distances between points translation, rotation and reflection? There are many sources which define rigid/isometric transformations as "transformations which preserve the distance between points", going on to say that "rotation, translation and (maybe) reflection are all types of rigid transformation".
What is not clear from this is whether these transformations are the only transformations which preserve pairwise distances, or whether there might be some more complex transformations which also do so.
Turning the question around: If I have a set of points, are translations, rotations and reflections the only transformation I can possibly apply if I wish to maintain distances between points?
 A: They're not the only ones: Consider $$ (x,y) \mapsto (x,-y) \mapsto (x+1,-y). $$ This is not a reflection about a line, nor a translation, nor a rotation.  However, it is a composition of a reflection and a translation.  The answer should be that translations, reflections, and rotations GENERATE the group of all isometries of the plane, i.e. every isometry is a composition of finitely many those.  (Auxiliary question: How many? Do we ever need more than two?)
("Glide reflection" is a term I've seen used for isometries like the one displayed above.  I don't know what degree of standardness that term has.)
In three dimensions, you could also rotate through a $1^\circ$ angle and then translate along the axis of rotation.  By contrast, in two dimensions, if you rotate and then translate, what you get is just a rotation about a different center.
A: I assume you talk about plane geometry.
Given a triangle $ABC$ and its image $A'B'C'$ under an isometry (so in particular these triangles are congruent), we can take $ABC$ to $A'B'C'$ as follows:
First, reflect at the bisector of $AA'$ to map $A\mapsto A'$ (while mapping $B\mapsto B_1$, $C\mapsto C_1$). 
Then reflect at the bisector of $B_1B'$ (which passes through $A'$!) to map $B\mapsto B'$ (while mapping $A'$ to itself and $C_1\mapsto C_2$).
If $C_2\ne C'$, reflect at $A'B'$ (which must take $C_2$ to $C'$). In the end we have transported $ABC$ to $A'B'C'$ i at most three reflections. As we can combine two reflections into a rotation or a translation, we have found an isometry of the desired type.
Now it only remains to show that this isometry works not only for these three points, but for the whole plane. But that is clear because the image $D'$ of any point $D$ is determined by the fact that $A'D'=AD$, $B'D'=BD$, $C'D'=CD$.
A: I assume you talk about vectorial isometries. 
In dimension $n$, given a unitary transformation, there always exists an orthonormal basis where the transformation is represented by a block-diagonal matrix of diagonal terms either +1, -1 or a 2 by 2 rotation matrix.
So:


*

*if $n=2$, any isometry is either a reflection or a rotation. 

*if $n=3$, it can be a reflection, a rotation or a combination of both 

A: Here is another answer to this old posting.

Definition: Suppose $(X,d)$ is a metric space. A function $F:X\rightarrow X$ such that
$$ d(F(x),F(y))=d(x,y),\qquad x,y\in X$$
is called a rigid transformation on $X$.

It is easy to check that the composition of rigid transformations is also a rigid transformation.
Suppose $(X,\langle\cdot,\cdot\rangle)$ is a real inner product space. The inner product induces a norm on $X$ given by $\|x\|:=\sqrt{\langle x,x\rangle}$, with in turn defines a distance $d:X\times X\rightarrow [0,\infty)$ given by
$$d(x,y)=\sqrt{\langle x-y,x-y\rangle}$$

*

*For any $b\in X$ a translation by $a$ is the map $T_a:X\rightarrow X$ defined as
$$T_a(x)=x-a$$

*A linear isometry on $X$ is a linear map $R:X\rightarrow X$ such that
$$\langle Rx,Ry\rangle =\langle x,y\rangle,\qquad x,y\in X$$
We denote the space of linear isometries by $O(X)$.

It is easy to check that both translations and linear isometries maps are rigid transformations. We have the following result

Theorem: Suppose $X$ is a real  inner product space. If $F$ is a rigid transformation, then there are unique $G\in O(X)$ such that $$F=T_b\circ G$$
where $b=-F(0)$.

Proof: Suppose $F$ is a rigid transformation and  $G=F-F(0)$. Clearly $G$ is a rigid transformation that fixes $0$, and $\|G(x)\|=\|x\|$ for all $x\in X$. We now show the $G$ is a linear isometry. Hence, for any $x,y\in  X$
$$\begin{align}
2\langle G(x),G(y)\rangle &= \|G(x)-G(y)\|^2-\|G(x)\|^2-\|G(y)\|^2\\
&=\|x-y\|^2-\|x\|^2-\|y\|^2=2\langle x,y\rangle,
\end{align}
$$
That is
$$ \langle G(x),G(y)\rangle = \langle x,y\rangle,\qquad x,y\in X$$
Finally, we show that $G$ is linear.
$$\begin{align}
\|G(x+\alpha y)-G(x)-\alpha G(y)\|^2&=\|G(x+\alpha y)-G(x)\|^2-2\alpha \langle G(x+\alpha y)-G(x),G(y)\rangle +\alpha^2 \|G(y)\|^2\\
&=\|x+\alpha y - x\|^2-2\alpha\langle G(x+\alpha y),G(y)\rangle+2\alpha\langle G(x),G(y)\rangle +\alpha^2\|y\|^2\\
&=2\alpha^2\|y\|^2-2\alpha\langle x+\alpha y,y\rangle +2\alpha\langle x,y\rangle\\
&=2\alpha^2\|y\|^2-2\alpha\langle x,y\rangle +2\alpha^2\|y\|^2 +2\alpha\langle x,y\rangle=0
\end{align}
$$
Therefore, $F=G+F(0)=T_{-F(0)}\circ G$, where $G\in O(X)$.
