Why does NMF of a symmetric matrix yield orthogonal matrices which are not transpose identical?

Consider the non-negative factorization of a positive, real symmetric matrix A. Non-negative factorization of this matrix yields A = WH where W != H^T.

Yes, there are special cases of A = WW^T factorization, but I understand that this is an imposed constraint. Thus, my question is two-fold:

1. Intuitively, why does NMF for a symmetric A not yield A = WW^T?
2. What is the meaning of non-transpose-identity in WH matrices when A is symmetric? How can the orthogonal factors in W and H be interpreted in terms of relationships in A? Is assignment of a loading to W over H a stochastic process? Thus, can conclusions be drawn about the relationships between columns across factors within W or H alone?

My application: I'm trying to decompose gene associations from a gene adjacency matrix. I want to not only decompose the largest sources of variation (i.e. SVD) but also orthogonal and nested gene associations. NMF is ideal for decomposing orthogonal associations as latent factors, and then a similarity generated from relationships across NMF latent factors could enable decomposition of first-order nested gene associations. This sets up an iterative framework for decomposition of higher-order nested gene associations. However, I worry that NMF will randomly partition genetic interaction information due to asymmetry of orthogonal matrices, thus resulting in exponential fragmentation of information with each iteration. I'm at a loss for how to prove or disprove this notion.

The reason why $$H\neq W^T$$ is that for two vector $$a$$ and $$b$$ to be orthogonal, you need:
$$\sum_{i=1}^n a_ib_i=0$$
This will typically be achieved by having $$a_ib_i$$ to be positive at some indices and negative at others - and this is not possible under non-negative matrix factorization. If you factor $$A$$ as $$WW^T$$ (maybe with eigendecomposition or Cholesky), the resulting $$W$$ will almost certainly have negative numbers in it.