I'm not sure if the formula below is useful, but at least it can be used to numerically count edge-invariant walks classes for small $n,l$. Let's assign a unique variable to each edge and each vertex of the complete graph, say, $x_{ij}$ to an edge $(i,j)$ and $y_i$ to a vertex $i$. Then, we assign to each edge $(i,j)$ the weight $x_{i,j}y_j$ and consider the graph adjacency matrix $A$. For example, for $n=4$ the matrix is $$A = \begin{bmatrix} x_{11}y_1 & x_{12}y_2 & x_{13}y_3 & x_{14}y_4\\ x_{21}y_1 & x_{22}y_2 & x_{23}y_3 & x_{24}y_4\\ x_{31}y_1 & x_{32}y_2 & x_{33}y_3 & x_{34}y_4\\ x_{41}y_1 & x_{42}y_2 & x_{43}y_3 & x_{44}y_4 \end{bmatrix}.$$ Then the number of classes of edge-invariant of length $l$ is given by the number of distinct monomial terms in $$[y_1,y_2,\dots,y_n]\cdot A^l\cdot [1,1,\dots,1]^T.$$ For example, for $n=2$ and $l=3$ we get the polynomial: $$x_{11}^3y_1^4 + x_{11}^2x_{12}y_1^3y_2 + x_{11}^2x_{21}y_1^3y_2 + 2x_{11}x_{12}x_{21}y_1^3y_2 + x_{11}x_{12}x_{21}y_1^2y_2^2 + x_{12}^2x_{21}y_1^2y_2^2 + x_{12}x_{21}^2y_1^2y_2^2 + x_{11}x_{12}x_{22}y_1^2y_2^2 + x_{11}x_{21}x_{22}y_1^2y_2^2 + x_{12}x_{21}x_{22}y_1^2y_2^2 + 2x_{12}x_{21}x_{22}y_1y_2^3 + x_{12}x_{22}^2y_1y_2^3 + x_{21}x_{22}^2y_1y_2^3 + x_{22}^3y_2^4,$$ where the monomials describe equivalence classes of walks (more specifically, $x$'s describe $e_p$ and $y$'s describe $f_p$) and the coefficients give the size of each class. There are $14$ distinct monomials here and only two of them have coefficient $2$, as expected. Here is my SAGE implementation of this formula: # generator for variables class VariableGenerator(object): def __init__(self, prefix): self.__prefix = prefix @cached_method def __getitem__(self, key): return SR.var("%s%s"%(self.__prefix,key)) def NumEIClasses2(n,l): x = VariableGenerator('x') y = VariableGenerator('y') R = PolynomialRing(QQ,[x[i] for i in range(n*n)] + [y[i] for i in range(n)]) A = matrix([[x[n*i + j]*y[j] for j in range(n)] for i in range(n)]) u = vector([1 for i in range(n)]) Y = vector([y[i] for i in range(n)]) P = R( (Y.row() * A^l * u.column())[0,0] ) #print P # print the resulting polynomial return P.hamming_weight() For example, for $n=4$ it gives counts 16, 64, 244, 856, 2728, 7892, 20876, 51020, 116408, ...