Fundamental lemma: why is the transfer factor a power of q Let $k$ be a finite field of sufficiently large characteristic, $F = k((t))$ and $\mathfrak{o} = k[[t]]$. Let $G$ be a reductive algebraic group defined over $\mathfrak{o}$. Roughly stated, for sake of brevity, the Lie algebra fundamental lemma of Ngô states there is an equality of the form
$$
\mathcal{O}^\kappa_a = q^r\mathcal{SO}_{a_H}
$$
where $\mathcal{O}^\kappa$ is a $\kappa$-orbital integral for $G$ and $\mathcal{SO}$ is a stable orbital integral for an endoscopic group $H$ associated to $\kappa$ (or more precisely and endoscopic datum). Here $\kappa$ is a character of a cohomology group associated to stable conjugacy class $a$ in $\mathrm{Lie}(G)(F)$, and $a_H$ is the corresponding conjugacy class in $\mathrm{Lie}(H)(F)$. A more detailed and precise formulation is in [1], but this should suffice for the question.
Consider transfer factor $q^r$ in front of the stable orbital integral: it is a simple power of $q$ and $r$ depends on $a$ and $a_H$ (the number $r$ is the difference in dimensions of the associated affine Springer fibers).

Is there a relatively simple, perhaps 'geometric' explanation or heuristic of why the transfer factor in the fundamental lemma is a simple power of $q$, as opposed to another polynomial in $q$ such as e.g. $q^{r_1} + 2q^{r_2}$ where $r_1$ and $r_2$ just depend on $a$ and $a_H$? 

Hales' paper [2] is a good relatively nontechnical summary of the fundamental lemma that briefly discusses this transfer factor but doesn't quite answer this question.
[1] Ngô, Bao Châu. Le lemme fondamental pour les algèbres de Lie. Publ. Math. Inst. Hautes Études Sci. No. 111 (2010), 1-169
[2] Hales, Thomas C. The fundamental lemma and the Hitchin fibration [after Ngô Bao Châu]. Séminaire Bourbaki: Vol. 2010/2011. Exposés 1027–1042. Astérisque No. 348 (2012), Exp. No. 1035, ix, 233-263.
 A: I was hoping an expert would have commented on this. In any case, from my understanding, it boils down to purity. One sees this by global considerations: look at the fiber over the Hitchin base $\mathcal A_G$; by the BBDG decomposition theorem the complex $f_*\bar{\mathbb Q}_\ell$ is a direct sum of simple perverse factors, and the purity theorem tells us that the eigenvalues of the Frobenius action of each summand is a power of $q$.
Ngô proves this 'by hand' for a 'good' open subset of the anisotropic locus, i.e., he verifies some special cases of this, then by a continuity argument is able to extend this over the entire locus.
Now the exact power of $q$, I believe, is coming from the choice of certain divisors denoted $\mathfrak D_H$ and $\mathfrak R^H_G$ related to the discriminant and resultants of the conjugacy class involved. But eventually the exponent measures the codimension of the endoscopic stratum $\mathcal A_H$ in $\mathcal A_G$, which locally is relates to the dimension of the affine springer fibers, as you know.
This is mentioned in part by the report by J.F. Dat and D.T. Ngô in Lemme Fondamental pour les Algèbres de Lie, pp. 20-21, but I haven't been able to track this down precisely in B.C. Ngô's original paper.
EDIT: Looking more closely, I'm not so certain that my first paragraph is the right answer. But if you feel so inclined, I believe the key is Lemme 8.5.7 in Ngô, where he makes the local calculations (point counting) for some simpler cases. If that is the case, then there doesn't quite seem to be a geometric reason for the power of $q$, except that it matches the expected form.
A: I am also not 100% sure, but my impression (if wrong, the error is mine)  from conversations with Hales was that the power-of-q part of the transfer factor is the easy part, simply the absolute value of a ratio of some  Weyl discriminant-style factors (which comes from the normalization of measures on the matching orbits).
This discriminant factor is written explicitly in the other excellent expository paper by Hales, https://arxiv.org/pdf/math/0312227.pdf (see formula (6.0.1)): the product over roots is what gives the power of q, simply because it is the absolute value. The ratio of volumes in the same formula 
I think is not part of the transfer factor, it's just a constant that can be removed by choosing the measures on G and T  (and is not a power of q).   
