About this problem ${a}$, I am wondering if there are 5 orbits in $A$? The 5 orbits separately contain elements which 3 are all the same, 2 of 3 are the same and all 3 are different?
I am confused about finding isotropy(stabilizer) group $G_a$, I am wondering is this correct? For example, isotropy group of $(1,1,1)$ is $(1)(2,3),(1)(2)(3)$. Is this correct?
About this problem $(b)$. Is there $3$ orbits which are one element sets, two element sets and three element sets?
Is it the same to find isotropy group in $(b)$?
Let $B=\{1,2,3\}$ then $S_3$ also act on $B$ and there is a relation with between avtoion on $B$ and $A$.
Let $G_{1},G_2,G_3$ are the stabilizers related the action on $B$ then,
$G_{(1,2,3)}=G_1\cap G_2 \cap G_3=e$ this olso show that orbit of the $(1,2,3)$ has six elements. You can also see that only $(1,1,1),(2,2,2),(3,3,3)$ have $3$ elements in their orbits as their stabilizer has two elements and rest of the elements has $6$ elements in thier orbits.
For $a)$ $|A|=27$ and it is easy to see that $(1,1,1),(2,2,2),(3,3,3)$ are in same orbit so rest of the orbit must have size $6$. $(27-3)/6=4$ so we have $5$ orbits in total.
For $b)$ we have three orbits, $$O_1=\{\{1\},\{2\},\{3\}\}$$ $$O_2=\{\{1,2\},\{2,3\},\{1,3\}\}$$ $$O_3=\{\{1,2,3\} \}$$