For which algebra $A$: $\mathrm{vect} \cong A-$mod In general there is an equivalence between a module category (see 
Module Category nLab) over a monoidal category $\mathcal{C}$ and $A-$mod, where $A$ is an algebra over $\mathcal{C}$:
$$\mathcal{M}_{\mathcal{C}} \cong A-mod$$
This is proven in Ostrik, Theorem 1 on page 10.
I now want to consider an explicit example. 
Let $k$ be a field. Let $\mathcal{C}=G-\mathrm{vect}$ be the category of $G$-graded vector spaces over $k$. And let $\mathcal{M}_{G-\mathrm{vect}}$ be the category $\mathrm{vect}_{k}$, all vector spaces over a field $k$. The module structure is given by the tensor product of vector spaces, where we use the forgetful functor on the $G$-graded vector spaces. 
Now I want to find such an algebra $A$ as in the statement above. In the proof of Theorem 1 in Ostrik, $A$ is given as $\underline{\mathrm{Hom}}(V, V)$ for an arbitrary vector space $V$. 
$\underline{\mathrm{Hom}}(V, V)$ is the representing object for the functor $X \mapsto \mathrm{Hom}(X \otimes V, V)$, with $X \in \mathcal{C}$=$G-\mathrm{vect}$. 
Unfortunately I have no idea how to find such a representing object, especially since $V$ is arbitrary. Does anyone have an idea on how I could proceed to get such a $V$ and $\underline{\mathrm{Hom}}(V, V)$?
 A: Let $\mathcal{M}$ be a module category over $\mathcal{C}$ given by the forgetful functor $F:\mathcal{C}\to \mathcal{M}=\textrm{Vect}$. Choose an arbitrary irreducible object $M\in\mathcal{M}$.
Now we compute the inner hom $\underline{Hom}(M,M)$. For every $V\in \mathcal{C}$ we should have 
$$Hom_\mathcal{C}(V,\underline{Hom}(M,M))\simeq Hom_\mathcal{M}(V\otimes M,M).$$
For every $g\in G$ let $V_g\in \mathcal{C}$ be a $1$-dim vector space graded by $g$. The we have $$Hom_\mathcal{C}(V_g,\underline{Hom}(M,M))\simeq Hom_\mathcal{M}(F(V_g)\otimes M,M)\simeq Hom_\mathcal{M}(M,M)=k,$$ hence $\underline{Hom}(M,M)$ has each $V_g$ exactly one time. Therefore $\underline{Hom}(M,M)\simeq \bigoplus\limits_{g\in G}{V_g}$. 
It is not hard to "guess" that the algebra structure given by composition is the group algebra structure on $kG$. Therefore $\underline{Hom}(M,M)$ is canonically isomorphic to $kG$. 
Also one can manually check that for every $kG$-module in $\mathcal{C}$ is free, hence the category of $kG$-modules in $\mathcal{C}$ is equivalent to $\textrm{Vect}$.
Hope that helps.
