Count Orbits and stabilizer Let $X$ be the set $\mathbb{Z}_9\times \mathbb{Z}_9$ and let $U_9$ denote the group of invertible elements in $\mathbb{Z}_9$. The group $G$ acts on $X$ defined by $u(x,y)=(ux,uy)$ where $u\in U_9$ and $x,y\in X$.
$(a)$ Count the elements in $U_9$
$(b)$ Speficy the set of fixed points $F(u)$ (I guess they mean the stabilizer?) for each $u\in U_9$.
$(c)$ Count the number of orbits $X$ under the action of $U_9$.

$(a)$ Every integer which are coprime to 9 are invertible in $\mathbb{Z}_9$, and so $U_9=\{1,2,4,5,7,8\}$
$(b)$ I'm not quite sure if I am doing it right here. Here is a theorem, the proof was left as an exercise in my textbook which I think I proved correctly, and I would like to use the result to solve this problem.
Theorem: Let $G$ be a group of permutations of $X$ and suppose that u belongs to $G(x\to y)$. Then
$$G(x\to y)=uG_x\text{ (where } G_x \text{ is the stabilizer)}$$
the left coset of $G_x$, with respect to u.
Since $u$ has an inverse I can rearrange the equation $F(u)=u^{-1}G(x\to y)$. In my case: $F(u)$ is the theorems $G_x$...
So let's take $F(1)$ as the first example. If what I am saying is right, then $F(1)=G(x\to y)$ and $|F(1)|=81$? That is, the elements of F(1) is $\{(0,0),(0,1),(0,2),(0,3),(0,4),...,(8,0),(8,1),...,(8,7),(8,8)\}$. This feels kinda wrong to me the more I think about it though. 
If i do the same for $F(2)$ I find that it's inverse is 5 (usually I make use of the euclidian algorithm to find the inverse but here I could easily see that 5 is the inverse of 2) and $|F(2)|=3*3=9$ and the elements are given by $F(2)=\{(0,0),(0,5),(0,1),(5,0),(5,5),(5,1),(1,0),(1,5),(1,1),\}$. Just in case I have misunderstood this totally, I will explain my calculations for some of the elements in $F(2)$.
I start off with $u^{-1}(x,y)=5(x,y)$. The first thing I do is to put $x=y=0$ and then I see $5(0,0)=(5*0,5*0)=(0,0)$ then i continue with $x=0,y=1$ which gives me $5(0,1)=(5,0)$ and lastly $5(0,2)=(0,10)\equiv (0,1) \text{ (mod 9)}$ and since I now see a repeating pattern, $x=0,y=1$ will be the same as $x=0, y=3$, I have found my first 3 elements for $F(2)$. I will see the same pattern for $x=1$ and $x=2$ and by this procedure I will find all 9 elements. 
The more I think about it the more unsure I become of this method. Is It correct to write it like $F(u)=u^{-1}G(x\to y)$? Or should I perhaps write it something like $G_u=hG(u\to y)$, where $h$ is the left coset. But how should I then specify the set of fixed points $F(u)$? How do i count it? I'm not even sure why I can see this as permutations. I thought permutations had to do with how you can arrange elements in a given set. Hope someone can help me. I haven't given so much though on $(c)$ yet so perhaps you shouldn't try to help me too much on that point. But I would be happy if you could help me out with $(b)$ and answer my questions. Thank you :)
 A: Part $(a)$ looks perfectly fine to me.  
$(b)$ Now, your action $U_9\curvearrowright\mathbb{Z}_9\times\mathbb{Z}_9$ is defined as:
$$u(x,y) = (ux,uy).$$
If you are looking for the collection of elements that each $u\in U_9$ fixes, then you need the collection of $(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9$ such that 
$$ux\equiv x\bmod{9}\quad\text{and}\quad uy\equiv y\bmod{9}.$$
That is, all pairs $(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9$ such that 
$$x(u-1)\equiv 0\bmod{9}\quad\text{and}\quad y(u-1)\equiv 0\bmod{9}.$$
So, using your notation:
$$F(1) = \{(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9\ |\ 0x\equiv 0y\equiv 0\bmod{9}\} = \mathbb{Z}_9\times\mathbb{Z}_9,$$
$$F(2) = \{(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9\ |\ 1x\equiv 1y\equiv 0\bmod{9}\} = \{0\}\times\{0\},$$
$$F(4) = \{(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9\ |\ 3x\equiv 3y\equiv 0\bmod{9}\} = \{(0,0),(3,3),(6,6),(3,6),(6,3),(0,3),(3,0),(0,6)(6,0)\},$$
$$F(5) = \{(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9\ |\ 4x\equiv 4y\equiv 0\bmod{9}\} = \{0\}\times\{0\},$$
$$F(7) = \{(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9\ |\ 6x\equiv 6y\equiv 0\bmod{9}\} = \{(0,0),(3,3),(6,6),(3,6),(6,3),(0,3),(3,0),(0,6)(6,0)\},$$
$$F(8) = \{(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9\ |\ 7x\equiv 7y\equiv 0\bmod{9}\} = \{0\}\times\{0\}.$$
and so on. fixed element $(x,y)\in\mathbb{Z}_9\times\mathbb{Z}_9$, the stabilizer is:
$$\text{Stab}(x,y) = \{u\in U_9\ |\ (ux,uy)=(x,y)\}.$$ 
Let me know if you need help with $(c)$. 
Edit: On a side note, if you are interested in computing the stabilizers, all you have to do is complete the list above for $F(5),\ F(7),\ F(8)$ and then, for any element of $\mathbb{Z}_9\times\mathbb{Z}_9$, say, $(3,6)$, you'll have:
$$\text{Stab}(3,6) = \{u\in U_9\ |\ (3,6)\in F(u)\}.$$
Pretty cool!
I've added the sets $F(5),\ F(7)$ and $F(8)$. Now, Burnside's Lemma says:
$$|X/G| = \frac{1}{|G|}\sum_{g\in G}|X^g|,$$
where $|X/G|$ is the number of orbits. The sets $X^g$ are precisely our $F(u)$'s. $|U_9|=6$, so:
$$|\mathbb{Z}_9\times\mathbb{Z}_9/U_9| = \frac{1}{6}\left(|F(1)|+|F(2)|+|F(4)|+|F(5)|+|F(7)|+|F(8)|\right).$$
