Is there a tensor product of $G$-sets? We can take the tensor product of two vector spaces, and the tensor product of two modules.  I'm wondering if the same can be done for group actions.
Let $G$ be a group which acts on two sets $X$ and $Y$.  My question is there a definition for the tensor product of $X$ and $Y$?  
If so, what kind of object will it be?  The tensor product of modules need not be a module, it can just be an abelian group.  Similarly, is it possible that the tensor product of $X$ and $Y$ can just be a group rather than another $G$-set?
 A: Very generally, if $A$ is a monoid object in a monoidal category, we can define a "tensor product over $A$" $X\otimes_A Y$ between a right $A$-module $X$ and a left $A$-module $Y$ to be a coequalizer of the two maps $X\otimes A\otimes Y\to X\otimes Y$ (where the first map uses the multiplication $X\otimes A\to X$ and the second uses the multiplication $A\otimes Y\to Y$).  A bit more generally, if the monoidal product preserves coequalizers in each variable, then we also have a tensor product of bimodules: if $X$ is an $(A,B)$-bimodule and $Y$ is a $(B,C)$-bimodule for three monoid objects $A,B,C$, then $X\otimes_B Y$ will naturally have the structure of an $(A,C)$-bimodule.  In particular, if we have not just a monoidal category but a symmetric monoidal category and $A$ is a commutative monoid object, $X\otimes_A Y$ has a natural $A$-module structure (by considering $X$ and $Y$ as $(A,A)$-bimodules).
For the category of abelian groups with the (symmetric) monoidal structure given by the tensor product of abelian groups, this general definition gives the usual definition of tensor products of modules over rings.
To get an analogous "tensor product of $G$-sets", we need to pick a monoidal structure on the category of sets.  The obvious choice is the categorical (cartesian) product.  This then gives the following definition: if $A$ is a monoid, the "product over $A$" of a right $A$-set $X$ and a left $A$-set $Y$ is the set $$X\times_A Y=X\times Y/\sim$$ where $\sim$ is the equivalence relation that identifies $(xa,y)$ with $(x,ay)$ for each $x\in X,y\in Y,a\in A$.  This is just a set in general, but when $A$ is commutative $X\times_A Y$ is naturally an $A$-set by the action of $A$ on either coordinate.
In the special case that $A$ is a group, left $A$-sets and right $A$-sets are the same (a left action gives a right action by $(x,a)\mapsto a^{-1}x$), so we can also define a product over $A$ of two (left) $A$-sets by $X\times_A Y=X\times Y/\sim$ where $\sim$ identifies $(ax,y)$ with $(x,a^{-1}y)$.  This will still just be a set and not an $A$-set in general, though.
