Pullback along Frobenius morphism

Consider the $p$-th tensor power $\mathcal{M}^{\otimes p}$. The group $\mathbb{Z}/p\mathbb{Z}$ acts by cyclic permutations on it. Denote its generator by $\sigma$. There is a map from coinvariants to invariants $$(\mathcal{M}^{\otimes p})_{\sigma}\xrightarrow{1+\sigma+\dots+ \sigma^{p-1}}(\mathcal{M}^{\otimes p})^{\sigma}$$ Its kernel is canonically isomorphic to $F^*\mathcal{M}$ via the map $s\mapsto s^{\otimes p}$. This is proven in Lemma 6.9 here:http://arxiv.org/pdf/1509.08784v1.pdf for the affine case and the general case follows by functoriality.

Edit: By the way, its cokernel is also isomorphic to $F^*\mathcal{M}$.

Using the notation from my paper on Adams operations for the Schur functors of hook type, there ought to be a long exact sequence

$$0 \to F^*(\mathcal{M}) \to L_1^q(\mathcal M) \to \dots \to L_{q-1}^q(\mathcal M) \to 0$$

that generalizes the short exact sequence that you wrote down for $q=2$; the corresponding equation in the Grothendieck group holds, as explained in the paper, and that motivates the guess, which I remember wondering about long ago.