Consider a smooth manifold $M$ (assume it boundary-free and orientable) and a tensor field $\mathcal{G}\in\Gamma(\otimes^hTM\otimes^kT^*M)$. Let $\Phi:\mathbb{T}^p\times M \rightarrow M$ be a torus action on the manifold. What is the definition of the average of the tensor $\mathcal{G}$ with respect to the action $\Phi$?
I think it should be defined as
$$ \bar{\mathcal{G}} = \frac{1}{\int_M d\alpha}\int_M (\Phi_{\alpha})^*\mathcal{G}\;d\alpha_1\wedge ... \wedge d\alpha_p. $$
Is my intuition correct? It seems at least to be coherent with the fact that when $\mathcal{G}$ is invariant with respect to the torus action, then $(\Phi_\alpha)^*\mathcal{G}=\mathcal{G}$ and hence $\bar{\mathcal{G}}=\mathcal{G}$ as expected.
You should probably average it with respect to the action (but your integral is over $M$). If $G \circlearrowright M$ is a smooth action, $G$ is compact Lie group, $\nu$ the Haar (probability) measure on $G$ and we also denote by $g$ the map $M \ni p \mapsto g\cdot p \in M$, it seems that $$\overline{\mathcal{G}}_x(\theta^1,\ldots, \theta^h, X_1,\ldots, X_k) = \int_{G} (g^*\mathcal{G})_x(\theta^1,\ldots, \theta^h, X_1,\ldots, X_k)\,{\rm d}\nu(g)$$makes more sense, where $x \in M$, $\theta^i \in T_x^*M$ and $X_j \in T_xM$ for all $1 \leq i \leq h$ and $1 \leq j \leq k$ are all fixed. Then we have that $g^*\overline{\mathcal{G}} = \overline{\mathcal{G}}$ for all $g \in G$, by construction.