= Solution
At a locked state choose the rotating frame with $\theta_i'=0$. For small phase differences, <Taylor expansion> gives $h(\theta_j-\theta_i)-h(0)=h'(0)(\theta_j-\theta_i)+O(|\theta_j-\theta_i|^2)$. Let the <graph Laplacian> have $L_{ii}=\sum_jd_{ij}$ and $L_{ij}=-d_{ij}$ for $i\ne j$. The leading locked equations are
$$
0=\xi-\Omega_1\mathbf1-h'(0)L\theta.
$$
If coupling is balanced, $\sum_jd_{ij}=\sum_jd_{ji}$ for every $i$, so $\mathbf1^{\mathsf T}L=0$. Summing over cells cancels all linear coupling terms and gives
$$
\boxed{\Omega_1=\frac1N\sum_{i=1}^N\xi_i+O(\max_{ij}|\theta_j-\theta_i|^2).}
$$
In particular symmetric reciprocal gap-junction conductances satisfy this condition. To leading order the collective physical <frequency> is $1+\epsilon N^{-1}\sum_i\xi_i$.
\b[The arithmetic-mean conclusion requires this extra balance assumption.] The printed condition $d_{ii}=0$ alone is insufficient. For a connected directed coupling network let $\pi^{\mathsf T}L=0$, normalized by $\pi^{\mathsf T}\mathbf1=1$. The valid <weighted collective oscillator frequency> result is
$$
\boxed{\Omega_1=\sum_i\pi_i\xi_i+O(\max|\theta_j-\theta_i|^2).}
$$
For two cells with $d_{12}=p>0$, $d_{21}=q>0$, one has $\pi=(q,p)/(p+q)$ and $\Omega_1=(q\xi_1+p\xi_2)/(p+q)$ at linear order. For example $p=1,q=2,\xi_1=0,\xi_2=\delta$ gives $\Omega_1=\delta/3$, rather than $\delta/2$, while the locked phase difference is $O(\delta)$ when $h'(0)\ne0$. This supplies a counterexample with arbitrarily small phase differences. It also identifies exactly why reciprocal coupling recovers the requested mean: the left <null space> is then spanned by the uniform <vector>.
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