moduli space of triangles I found an article which seems to be aimed for general audience. I couldn't understand sentences about triangles. The link to the article is the following.
http://www.abelprize.no/c57681/artikkel/vis.html?tid=57787.
The author mentions the moduli space of triangles, up to similarity. I interpret symmetries there as isometries of $\mathbb{R^2}$. Non-trivial symmetries of equilateral triangles over the circle seems to be the 1/3-rotation around the center of the triangle.
Now I don't understand the sentence The image of the family into the universal family is constant, since the triangle is the same for all points in the circle. It seems to me that triangles are not the same. I thought the universal family is the space of triangles before equivalence. Am I misunderstanding what universal family is? And what does it mean to say that the family doesn't contain some other family?
I have a trouble in understanding moduli spaces and this simple-looking example confuses me even more.
 A: 
I thought the universal family is the space of triangles before equivalence.

No, the universal family is supposed to be a family (over the moduli space) which contains every isomorphism (i.e. similarity) class exactly once. More precisely, if $M$ is the moduli space and $U$ is the universal family, there is a map
$$ \pi: U \rightarrow M$$
such that for a point $p \in M$, the fibre $\pi^{-1}(p)$ is a triangle in the isomorphism class $p$. 
Up to similarity there is a unique equilateral triangle, so in the moduli space $M$ there is a unique point $p_{eq}$ which is the isomorphism class of equilateral triangles. So if the universal family exists, then the fibre $ \pi^{-1}(p_{eq})$ should be an actual equilateral triangle. 

And what does it mean to say that the family doesn't contain some other family?

To say that the universal family $\pi: U \rightarrow M$ contains every other family means that, for any family of triangles $: \phi: T \rightarrow Y$, there is a map $f: Y \rightarrow M$ such that the pullback family $U \times_f Y \rightarrow Y$ is isomorphic to $ T \rightarrow Y$. If you don't know about pullbacks don't worry the important point is that this would imply that for any $y \in Y$, the triangle $\phi^{-1}(y)$ is isomorphic to the triangle $\pi^{-1}(f(y))$. In particular if all the triangles in the family $T$ are isomorphic (as in the example in the link) then we would have to have $f: Y \rightarrow M$ being a constant map. But pulling back via a constant map always gives a trivial family $\Delta \times Y$ (for some fixed triangle \Delta)$. In the example the "twisted" family over the circle is not trivial, and that contradicts the existence of the universal family.
