Invariance of a bilinear (in generators) under a subgroup of a Lie group Let $L^a$ be generators of a Lie algebra of a compact connected Lie group $G$ in some irrep ($a$ indexes the generators). Let $\phi_k^a$ be the standard exponential coordinates for a group element $k$. Consider the following sum of group elements in any subgroup $K\subseteq G$ (sums over $a,b$ implied):
$$
\sum_{k\in K} \exp{(\phi^a_k L^a)}
$$
This sum is proportional to the projection onto any trivial irreps of $K$ in the irrep of $G$, so it is invariant under any conjugation by $r \in K$.
Let's expand it in the generators, obtaining the following quadratic leading-order term (the linear term cancels since for any $k\in K$, the inverse is also in the sum):
$$
\left(\sum_{k\in K}\phi_{k}^{a}\phi_{k}^{b}\right)L^{a}L^{b}\equiv f^{ab}L^{a}L^{b}
$$
I conjecture that the above quantity is also invariant under conjugation by $r \in K$.
Note that the matrix $f$ is a sum over outer products of the coordinates of the subgroup elements, acting on the space of the adjoint irrep of $G$. Under conjugation by $r\in K$, the generators transform as $L^a \to R^{ab} L^b$, where $R=\text{ad}(r)$. So what I think is true is that the matrix $f$ is invariant under conjugation by such rotations.
I've shown this for all subgroups of $K\subset SO_3$, discrete and continuous. If the adjoint irrep of $SO_3$ branches to only one irrep of $K$, it's simple: $f^{ab}\propto\delta^{ab}$. For $K=Z_N$ (rotations around the $z$-axis by multiples of $2\pi/N$), $f \propto \text{diag}(0,0,1)$, which makes sense since the $\phi^a$ are proportional to the axis of rotation. I have neither found a counter-example for other groups nor have I proven this more generally. I suspect the sum can also be over a class instead of a subgroup.
 A: One way around the problem of non-unique coordinates in the expansion is to manually average out the group element over the group before expanding. For each $k\in K$, pick a particular coordinate to represent it; let's say we use one for which $\phi^a_k \phi_k^a$ is minimized. Then, on the group level, use invariance of the sum under conjugation by $h\in K$ to write:
$$
\sum_{k\in K}e^{\phi_{k}^{a}L^{a}}=\frac{1}{2}\left(\sum_{k\in K}e^{\phi_{k}^{a}L^{a}}+\sum_{k\in K}e^{\phi_{h^{-1}kh}^{a}L^{a}}\right)=\cdots=\frac{1}{|K|}\sum_{k,h\in K}e^{\phi_{h^{-1}kh}^{a}L^{a}}
$$
Expanding the above sum then yields a second order term that is invariant basically by definition. Letting $R_k=\text{ad}(k)$ and shortening $R^{ab}\phi^b \equiv R\phi$, the relevant part of the second-order term reads
$$
\sum_{k,h\in K}\left(R_{k}\phi_{h}\right)^{a}\left(R_{k}\phi_{h}\right)^{b}L^{a}L^{b}
$$
Conjugating by $r\in K$, we can then use the finite group re-summation property to check the invariance:
$$
\sum_{k,h\in K}\left(R_{r}R_{k}\phi_{h}\right)^{a}\left(R_{r}R_{k}\phi_{h}\right)^{b}=\sum_{k,h\in K}\left(R_{rk}\phi_{h}\right)^{a}\left(R_{rk}\phi_{h}\right)^{b}=\sum_{k,h\in K}\left(R_{k}\phi_{h}\right)^{a}\left(R_{k}\phi_{h}\right)^{b}
$$
Same thing can be done on the level of classes.
