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Let $\Gamma$ be a group and denote by $\underline{\Gamma}$ the tensor category whose objects are simply the group elements, hom-sets only contain identities and the tensor product is given by the group product. The following definitions are standard:

Let $\mathcal{C}$ be a category. An action of $\Gamma$ on $\mathcal{C}$ is a monoidal functor $\underline{\Gamma}\rightarrow \text{Aut}(\mathcal{C})$.

In a similar fashion one has the following definition.

Let $\mathcal{C}$ be a monoidal category. An action of $\Gamma$ on $\mathcal{C}$ is a monoidal functor $\underline{\Gamma}\rightarrow \text{Aut}_{\otimes}(\mathcal{C})$. Here $\text{Aut}_{\otimes}(\mathcal{C})$ is the category of tensor auto-equivalences.

In the book tensor categories by Etingof, Gelaki, Nikshych and Ostrik, there is the following exercise:

Show that any action of $\Gamma$ on $\text{Vec}$ corresponds to an element in $H^2(\Gamma, k^*)$. Show that any action of $\Gamma$ on $\text{Vec}$ viewed as a monoidal category is trivial.

I was wondering why any action of $\Gamma$ on $\text{Vec}$ is trivial when we view the latter as a monoidal category. The only difference with the first part of the question is that we only allow tensor-auto-equivalences. How many tensor-auto-equivalences are there on $\text{Vec}$?

EDIT

For those who are wondering how to prove the first part of the exercise, I'll include the details in this post.

Suppose $\mathcal{C}=\ _k\text{Vec}$ which we view as an abelian category. That means that $\text{Aut}(\mathcal{C})$ are those additive functors that are auto-equivalences.

Let $\Gamma$ act on $\mathcal{C}$. Then for any $\gamma\in \Gamma$ we have an auto-equivalence $T_{\gamma}$. Since $T_{\gamma}$ is an auto-equivalence, it takes any one dimensional space to a one dimensional space. Moreover, there are natural isomorphisms $\theta_{\gamma_1,\gamma_2}:T_{\gamma_1}\circ T_{\gamma_2}\rightarrow T_{\gamma_1\gamma_2}$. This natural isomorphism is completely determined by it's action on the one-dimensional space $k$, the corresponding isomorphism is given by an invertible number which I will also denote by $\theta_{\gamma_1,\gamma_2}$.

Since $T$ is a monoidal functor and since $_k\text{Vec}$ is strict, we have that $$\theta_{\gamma_1\gamma_2,\gamma_3}\theta_{\gamma_1,\gamma_2}=\theta_{\gamma_1,\gamma_2\gamma_3}\theta_{\gamma_2,\gamma_3}.$$ This says that $\theta_{-,-}\in Z^2(\Gamma,k^*)$.

I guess one says two actions $T_1,T_2:\underline{\Gamma}\rightarrow \text{Aut}(\mathcal{C})$ are the same it the functors $T_1,T_2$ are monoidally equivalent. One should (probably) find the following statement. The functors $T_1,T_2$ are monoidally equivalent if and only if the corresponding $2$-cocycles are cohomologous. Carefully writing down the definitions should give the proof (I didn't check this though).

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