This is for a part of problem that my professor wants me to do as extra credit, so I'd prefer guidance rather than an explicit answer. The title also didnt fit exactly what my problem was so I'll state it better now.
Let $\ell$ be a prime and $(\mathbb{Z}/\ell^n\mathbb{Z})^{\times}$ act on $(\mathbb{Z}/\ell^n\mathbb{Z})^{2}$ diagonally. For an orbit $(\mathbb{Z}/\ell^n\mathbb{Z})^{\times} \cdot\alpha$ with representative $\alpha$ such that $ord(\alpha) = \ell^k$ show that $\mid(\mathbb{Z}/\ell^n\mathbb{Z})^{\times}\cdot\alpha\mid = \phi(\ell^k)$ where $\phi$ is the Euler totient function.
I've realized that using the Orbit-Stabilizer theorem we are done if we show that $\mid Stab(\alpha)\mid = \frac{\phi(\ell^n)}{\phi(\ell^k)}$. I thought the best way to move forward might to be construct a surjective homomorphism $\psi:(\mathbb{Z}/\ell^n\mathbb{Z})^{\times} \to Stab(\alpha)$ with $ker(\psi) = (\mathbb{Z}/\ell^k\mathbb{Z})^{\times}$. It seems relatively easy to construct a surjective map, but then I have trouble showing its a homomorphism. Or to construct a homomorphism, but then I have trouble showing its surjective. Not to mention that the kernel has to be right as well. Im also not seeing a very natural homomorphism, maybe I'm just tired. The last thing I tried was the mapping
$$\sigma \mapsto \begin{cases} \sigma & \sigma\cdot\alpha = \alpha \\ 1 & \sigma\cdot\alpha \not = \alpha \end{cases}$$ which was essentially a last ditch effort. It wasnt hard to show the top part followed the definition of a homomorphism, but the bottom doesnt work as if we take $\ell = 2$, $n =2$, $\sigma = 3$ and $\alpha = (1,3)$ then $3^2(1,3) = (1,3)$. I've tried a few other things that were all dead ends for me. I just need like an inch forward I think and I can get the rest.
Hints. I think you should compute the cardinality of the orbit directly.
Preliminary step: if $a\in\mathbb{Z}/\ell^n\mathbb{Z}$ is not the zero class, show that $a=u \bar{\ell}^{n-k}$, where $u$ is invertible and $1\leq k\leq n$. Compute $ord(a)$ in terms of $k$.
Then, convince yourself that if $ord(\alpha)=\ell^k$, then $\alpha=(\bar{\ell}^{n-k}a, \bar{\ell}^{n-k}b)$ where at least one of $a$ or $b$ lies in $(\mathbb{Z}/\ell^n\mathbb{Z})^\times$, say $a$, so that $b=a+\cdots+a$ $m$ times, for some $m$ (why ?)
Deduce carefully that the orbit of $\alpha$ has the same number of elements as the orbit of $\bar{\ell}^{n-k}$ (for the action of $(\mathbb{Z}/\ell^n\mathbb{Z})^\times$ on $\mathbb{Z}/\ell^n\mathbb{Z}$).
Use the first step to conclude that this last orbit is exactly the set of elements of order ??? to get the formula you seek.