Let $k$ be a perfect field of characteristic $p>0$, denote by $W=W(k)$ the associated ring of ($p-$)Witt vectors and consider the constant group scheme $G=(\mathbb Z/p^n\mathbb Z)_k$ over $k$. I would like to understand its contravariant Dieudonné module $M(G)$.
The underlying $W(k)$-module structure of $M(G)$ is easily seen to be $W(k)/(p^n)$, as the base ring $W(k)$ is a PID and by functoriality using that multiplication by $p^l$ in $G$ is $0$ only when $l\geq n$.
By étaleness of $G$ or checking the definitions directly, the relative Frobenius $F_{G/k}:G\rightarrow G^{(p)}$ is an isomorphism, hence the Frobenius action on $M(G)$ must be a $W(k)$-semi linear automorphism. Fixing a basis element $e$ for $M(G)$, we must have $F(e)=\alpha e$ for some $\alpha \in W(k)^{\times}$.
Is it possible to rescale the basis element $e$ so that we may assume $\alpha = 1$ ?
If I rescale $e$ by $e':=\lambda e$ for some $\lambda \in W(k)^{\times}$, I see that $\alpha$ is transformed into $\alpha':=\frac{\sigma(\lambda)}{\lambda}\alpha$ where $\sigma$ is the Frobenius automorphism of $W(k)$. The question is whether I can choose $\lambda$ so that $\alpha' = 1$.
This problem made me think of Hilbert's 90 theorem as stated here Corollary $1$ page $2$, but I am unsure if it can be applied to this context. Sure, it would work if $k$ was finite of cardinal $q=p^f$, as $L:=\operatorname{Frac}(W(k))$ would then be the unramified extension of $\mathbb Q_p$ of degree $f$, which is cyclic with Galois group generated by $\sigma$. How about a general $k$ ?
EDIT: I understand that my question can be solved by duality. In fact, the Dieudonné module $M(H)$ of $H=\mu_{p^n}$ the group-scheme of $p^n$-roots of unity over $k$ is easier to compute. An integer $m$ acts on the points of $H$ by raising an element to its $m$-th power. Thus yet again, we see that multiplication by $p^l$ in $H$ is trivial if and only if $l\geq n$. From this, we understand that the underlying $W(k)$-module structure on $M(H)$ is $W(k)/(p^n)$.
Now, the relative Frobenius $F_{H/k}$ is no other than $[p]$ in this case, whence by functoriality the Frobenius action $F$ on $M(H)$ is multiplication by $p$. It follows that the Verschiebung morphism is the identity.
On the other hand, $G$ and $H$ are Cartier duals of eachother. By general theory of Dieudonné modules, we deduce the relation $$D(G)=\operatorname{Hom}_{W(k)}(M(H),W(k)[1/p]/W(k))\cong W(k)/(p^n)$$ where $W(k)[1/p]$ is the fraction field of $W(k)$, and its quotient by $W(k)$ is the so-called "ring of Witt covectors" (cf. Oda's "The first de Rham cohomology group and Dieudonné module" section $3$). The isomorphism on the right is obtained by sending a homomorphism $\alpha$ to the class of $p^n x$, where $x\in W(k)[1/p]$ is a lift of $\alpha(1)$.
By duality, the Frobenius and Verschiebung on $D(G)$ act on a homomorphism $\alpha$ by $$(F\alpha)(x)=\alpha(Vx)^{\sigma}=\alpha(x)^{\sigma}$$ $$(V\alpha)(x)=\alpha(Fx)^{\sigma^{-1}}=\alpha(px)^{\sigma^{-1}}=p\alpha(x)^{\sigma^{-1}}$$ Under the above isomorphism taking the image of $1$, I deduce that for $z\in M(G)\cong W(k)/(p^n)$, I have $$Fz = \sigma(z) \quad \quad Vz = p\sigma^{-1}(z)$$
This last identities seem to confirm that my question above is correct: we can rescale $e$ to make the constant equal $1$. Whether it can be proved directly without duality with Hilbert's 90 theorem is still unclear to me.