The idempotent completion of a linear category $ \mathcal{C} $ may be though of in two different ways:

 - As a category with objects $ (X,e_X) $ where $ X $ is in $ \mathcal{C} $ and $ e_X $ is an idempotent in $ \text{End}(X) $. 
 - As the the additive subcategory of $ \text{Fun}^{\text{op}}(\mathcal{C}, \text{Vect}) $ generated by direct summands of $ \mathbb{Y}(X) $ for $ X $ in $ \mathcal{C} $ (where $ \mathbb{Y} $ is the Yoneda embedding).

To an object-idempotent pair $ (X,e_X) $ one can associate the (contravariant) functor 

$$ \begin{align} (X,e_X)^\sharp \colon \mathcal{C} &\to \text{Vect} \\ Y &\mapsto \{ f \in \text{Hom}(Y,X) \mid e_X \circ f = f \}  \end{align} $$

which provides an equivalence between these two categories.  

## Question ##

From the above discussion we see that the following are equivalent 

 - $ \mathbb{Y}\colon \mathcal{C} \to \text{Fun}^{\text{op}}(\mathcal{C}, \text{Vect}) $ is an idempotent completion.
 - Every contravariant functor from $\mathcal{C}$ to $\text{Vect}$ is naturally isomorphic to $ (X,e_X)^\sharp $ for some $X$ in $\mathcal{C}$ and some idempotent $e_X$ in $\text{End}(\mathcal{C}) $.

Under what assumption on $ \mathcal{C} $ do these statements hold? For example, they clearly hold if $ \mathcal{C} $ is semisimple. In particular I would like to know if they hold when $\text{Fun}^{\text{op}}(\mathcal{C}, \text{Vect})$ is semisimple (even if $ \mathcal{C} $ isn't).