The problem below was asked as a question here, but my questions are different.
The symmetric group $S_3$ operates on two sets $U$ and $V$ of order $3$.
Decompose the product set $U × V$ into orbits for the “diagonal action” $g(u,v) = (gu,gv)$ when
$(a)$ The operations on $U$ and $ V$ are transitive
$ (b) $ The operation on $U$ is transitive and the orbits for the operation on $V$ are $\{v_1\}$ and $\{v_2,v_3\}$.
For $(a)$, I wrote down the "multiplication table" (i.e., applied all 6 permutations to all elements in $U\times V=\{(a,b)| a,b=1,2,3\}$. It turned out that the orbits or $(1,1),(2,2),(3,3)$ are the same and are equal to $O_1=\{(1,1),(2,2),(3,3)\}$. Also, the orbits of all other elements are the same and are equal to $\{(1,2),(2,1),(3,2),(2,3),(1,3),(3,1)\}$. So the decomposition is $U\times V = O_1\cup O_2$.
What I have no idea about is where I used that the action is transitive. That's the first question.
For $(b)$, I used the description of the action of $S_3$ on $U\times V$ from the answer to the question referred to above, i.e. $e, (123), (132)$ fix both $2$ and $3$ (and hence $1$), and the transpositions swap $2$ and $3$. What I got is $5$ orbits (again by writing down the "multiplication table" and by inspecting it): $\{(1,1)\}, \{(1,2),(1,3)\}, \{(2,1),(3,1)\}, \{(2,2), (3,3)\}, \{(2,3),(3,2)\}$.
Is that correct? Also, I don't understand why $S_3$ acts on $U\times V$ as described (even though it has been asked (and answered) in the comments to the answer referred to above).