What are the group objects in the category of finite sets and bijections, and its functor category?

Question

An object $G$ in a category $\mathcal{C}$ is called a group object if, given any object $X$ in $\mathcal{C}$, there is a group structure on the morphisms $\operatorname{hom}\left(X,G\right)$ such that $X\mapsto \operatorname{hom}\left(X,G\right)$ is a (contravariant) functor from $\mathcal{C}$ to $\text{Grp}$.

Group objects of the category $\text{Set}$ are groups in the usual sense. Similarly, group objects of the category $\text{FinSet}$ are finite groups.

Assuming that good (non-tautological) descriptions exist,

1. What are the group objects of the category $\text{FinBij}$, the category whose objects are finite sets, and whose morphisms are bijections?

2. What are the group objects of the functor category of $\text{FinBij}$, the category whose objects are functors of $\text{FinBij}$, and whose morphisms are natural transformations?

Answer 1

The question uses a nonstandard definition of group object.

Under the ordinary definition of group object, the answer to the first question is

1. For the identity morphism $1 \to G$ and the multiplication morphism $G \times G \to G$ (if products exist...) to live in a category whose morphisms are bijections, $G$ must have cardinality $1$.

The definition in the question is equivalent to the usual only when the category has finite products. Although your definition would seem to be more general, in a category with terminal object, the subcategory generated by that object has a unique structure of finite products (all of which are equal to the terminal object) and one would therefore expect that the terminal object fits any definition of "group object". As written in Mariano's answer this is not the case for your definition. Therefore a different phrase than "group object" should be used for this more general definition.

Answer 2

There is no group object: since you only have bijections in your category, for all G there exists an X such that hom(x,G) is empty, so not a group in any way at all.