One thing that got me confused at the start was the phrase "labeled
pentagons".  I assumed that it would be a pentagon with labels at the
vertices, so that one could have a pentagon P1 with labels 01234 along
the perimeter, another pentagon P2 with labels 43210, and another P3 with
01432.  But later on you basically say that the vertices are numbered 0--4 *in circular order*
in either sense, so P1 and P2 are OK but P3 is not. 

P1 and P2 would not be congruent (since mirroring is not
allowed) but P1 can be continuously deformed into P2. P1 cannot be deformed into P3 
in one sense, but in the Hausdorff metric there is a connected path
between them that goes through a degenerate state.

If one allows arbitrary numbering of vertices with labels 0--4, then
\CL_5 would have 12 connected components. Some pairs of components would
intersect at the degenerate polygons with one zero angle, along
circles. Circles will intersect at points correspsonding to doubly-deenerate 
pentagons


On fig?? It may be worth extending the diagonal line to show the pentagram.

One of the basic results of plane geometry is that an eequilqteral triangle has only one shape.

On the other hand. an equilateral quadrilateral can have many shapes,
ranging from a square or rhombus (diamond, lozenge), to some
degenrate shapes where two segments connected by one vertex, each
traveersed twice, and a single segment traversed back and forth four
times.