Let $$\begin{align*} r_{\theta} = \begin{pmatrix} \cos\theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \end{align*}$$ for $\theta \in[0,2\pi]$ . Then all $r_{\theta}$ forms $SO(2)$ .
Let $$\begin{align*} \phi_{n}:& SO(2) \to \text{Aut}(\mathbb{R}^{2}) \\ &r_{\theta} \to r_{n\theta} \end{align*}$$ be a homomorphism, and use it to construct semi-direct product $\mathbb{R}^{2} \rtimes_{\phi_{n}} SO(2)$.
Thus the operation in $\mathbb{R}^{2} \rtimes_{\phi_{n}} SO(2)$ will be $$\begin{align*}
(v, r_{\theta}) (u ,r_{\theta'}) &= (v+ r_{\theta}^{n}(u),r_{\theta + \theta '})
\\ &= (v+ r_{n\theta}^{}(u),r_{\theta + \theta '})
\end{align*}$$
I want to show that for different $n ,m \in \mathbb{N}$ , we have $$\begin{align*}
\mathbb{R}^{2} \rtimes_{\phi_{n}} SO(2) \not \cong \mathbb{R}^{2} \rtimes_{\phi_{m}} SO(2)
\end{align*}$$
I tried to show this by contradiction. Suppose there exists an isomorphism $\psi$, then it should preserve the order of the elements. I want to use this fact to construct a contradiction, but I got stuck for a quite long time. Any help on this? Thanks!
Define
$$\color{purple}{G(n)}:= \mathbb{R}^2\rtimes_{\varphi_n}SO(2)$$
with product defined relative to $\varphi_n$ as above. Consider integer powers of an element in $G(n)$:
$\color{purple}{(u,r_{\theta})}$
$(u,r_{\theta})^2 = (u,r_{\theta})(u,r_{\theta}) = (u+r_{n\theta}(u),r_{2\theta})$
$(u,r_{\theta})^3 = (u,r_{\theta})(u,r_{\theta})^2 = (u,r_{\theta})(u+r_{n\theta}(u),r_{2\theta}) = \big(u+r_{n\theta}(u+r_{n\theta}(u)),r_{3\theta}\big)$
$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }= (u+r_{n\theta}(u)+r_{n\theta}(r_{n\theta}(u)),r_{3\theta}) = (u+r_{n\theta}(u)+r_{2n\theta}(u),r_{3\theta})$
$\vdots$
$\color{blue}{(u,r_{\theta})^x = \bigg(\sum\limits_{k=0}^{x-1} r_{kn\theta}(u),r_{x\theta}\bigg)},$
where Linearity of the $r_{}$'s has been applied, as well as the special fact $r_{\alpha}r_{\beta} = r_{\alpha+\beta}$.
If $x$ is the order of $(u,r_{\theta})$, then: $$(u,r_{\theta})^x = \color{purple}{1_{G(n)}} = (0,r_0)$$ $$\implies \sum\limits_{k=0}^{x-1} r_{kn\theta}(u) = 0\text{ }(\in\mathbb{R}^2)\text{ }\text{ }\text{ and }\text{ }\text{ }r_{x\theta} = r_0\text{ }(\in SO(2)).$$ The latter of which gives: $$\theta = \frac{2\pi}{x}\color{purple}{l},\text{ for }l\in\mathbb{Z}.$$ Substituting this into the prior gives the master equation that a finitely ordered element of $G(n)$ satisfies:
$$\color{darkblue}{\sum\limits_{k=0}^{x-1}(r_{\frac{2\pi k}{x}})^{(nl)}(u) = 0}$$
Suppose now, that $\color{purple}{\Phi:G(n)\to G(m)}$ is an isomorphism and that $\color{purple}{(v,r_{\phi})} := \Phi((u,r_{\theta}))$. Since $\color{red}{\text{isomorphisms preserve order}}$, we have corresponding information for $(v,r_{\phi})$. That is: $$\phi = \frac{2\pi}{x}\color{purple}{l'},\text{ for }l'\in \mathbb{Z}\text{ and }$$
$$\color{darkblue}{\sum\limits_{k=0}^{x-1}(r_{\frac{2\pi k}{x}})^{(ml')}(v) = 0}.$$
Equating across the zero vector and combining the sums gives: $$\sum\limits_{k=0}^{x-1}\bigg[\big(r_{\frac{2\pi k}{x}}\big)^{(ml')}(v) - \big(r_{\frac{2\pi k}{x}}\big)^{(nl)}(u)\bigg] = 0$$ $$\implies \sum\limits_{k=0}^{x-1}\big(r_{\frac{2\pi k}{x}}\big)^{(ml')}\bigg[(v) - \big(r_{\frac{2\pi k}{x}}\big)^{(-ml')}\big(r_{\frac{2\pi k}{x}}\big)^{(nl)}(u)\bigg] = 0.$$ Since we know $0\notin SO(2)$, it must be the case that: $$\forall k\in \{0,...,x-1\}:\text{ }v = (r_{\frac{2\pi}{x}})^{(nl-ml')}(u).$$ Clearly $v$ can't be several different images of $u$ at once, so it must be the case that: $$nl = ml'.$$ But this important result gives:
$$v = u\text{ }\text{ and }\text{ }\phi = \frac{n}{m}\theta.$$ In other words, elements of finite order have designated (isomorphic) images of the form:
$$\color{darkorange}{\Phi\big((u,r_{\theta})\big) = \big(u,r_{(n/m)\theta}\big).}$$
Now, we may show $$\exists g\neq 1_{G(n)}\in ker(\Phi:G(n)\to G(m)),$$ which contradicts injectivity unless $m=n$. [Exercise] $\blacksquare$