Probably I'm having a fundamental misunderstanding of the problem ...

... but I think with just one more color it's trivial!

Consider an incidence matrix (Node x Node) with the colors given in the cells (here RGBYCM...)

Now a solution for N colors (instead of N-1) is straightforward, we just rotate the N colors left for each row.

NB:

• The colors must be symmetric to the diagonal, because it's not a directed graph - i.e. (a,b)=(b,a) etc. That's guarantied by the rotation.
• The diagonal must be "emptied" at the end, because circular edges aren't allowed- i.e. (a,a)=undef Hence one color is missing in each row and column.

    
N=3

   a  b  c
a     R  G
b  R     B
c  G  B
[download]

    
N=4

   a  b  c  d
a     R  G  B
b  R     B  Y
c  G  B     R  
d  B  Y  R
[download]

    
N=5

   a  b  c  d  e  
a     R  G  B  Y  
b  R     B  Y  C  
c  G  B     C  R        
d  B  Y  C     G    
e  Y  C  R  G
[download]

    
N=6

   a  b  c  d  e  f
a     R  G  B  Y  C
b  R     B  Y  C  M
c  G  B     C  M  R      
d  B  Y  C     R  G    
e  Y  C  M  R     B     
f  C  M  R  G  B
[download]