Isometries of $\mathbb{R}^2$ have infinitely many subgroups isomorphic to $O_2(\mathbb{R})$ .

Question

I have shown that for every $x ∈ ℝ^2$, the stabiliser of $x$ is a subgroup of $I_2(\mathbb{R})$ (the group of isometries on $\mathbb{R}^2$) that is conjugated with the orthogonal group $O_2(\mathbb{R})$. (You can take the translation $\tau_x : a ↦ a - x$ and show that the stabiliser of $x$ equals $\tau_x O_2(\mathbb{R})\tau_x^{-1}$.) I'm subsequently asked to conclude that $I_2(\mathbb{R})$ therefore has infinitely many subgroups isomorphic to $O_2(\mathbb{R})$. I assume the idea is that there are infinitely many stabilisers, and that they are somehow isomorphic to $O_2(\mathbb{R})$. However, $O_2(\mathbb{R})$ is, to my knowledge, not a normal subgroup of $I_2(\mathbb{R})$, so we don't have $\tau_xO_2(\mathbb{R})\tau_x^{-1} = O_2(\mathbb{R})$. Is it possible to construct an isomorphism between them? Or is my assumption wrong?