An n-sphere (unit) is the simplest and most symmetrical manifold described by a function $f:\mathbb{R}^n\to\mathbb{R}^1$ that is $$ \{x\in\mathbb{R}^n\,:\, f(x) = 0\} = \{x\in\mathbb{R}^n\,:\, x^\top x - 1 = 0 $$ where $f(x)=x^\top x - 1$.
What is the equivalent (simplest, most symmetric) manifold resulting from a function $f:\mathbb{R}^n\to\mathbb{R}^2$?
If $n, n - p \geq 1$, for the map $$\Phi: \Bbb R^{n + 2} \cong \Bbb R^{p + 1} \times \Bbb R^{n - p + 1} \to \Bbb R^2$$ defined by $$\Phi({\bf x}, {\bf y}) = ({\bf x}^\top {\bf x}, {\bf y}^\top {\bf y}) ,$$ the level set $L := \Phi^{-1}(1, 1)$ is the product $$S^p \times S^{n - p}$$ of spheres, whose isometry group has dimension $$\frac{1}{2} [n^2 - (2 p - 1) n + p^2)].$$ In the case $p = 1$, this formula specializes to $$\dim \operatorname{Iso}(S^1 \times S^{n - 1}) = {{n - 1} \choose 2} + 1,$$ but this dimension is maximal in the following sense.
First, we may as well assume that our defining function $\Phi: \Bbb R^{n + 2} \to \Bbb R^2$ has (constant) rank $2$ on level set $M$ of interest, so that $M$ is a smooth manifold of dimension $n$. It's (well) known that the maximal isometry group of an $n$-manifold is ${n \choose 2} = \frac{1}{2} n (n - 1)$, and that equality holds iff $L$ is isometric to a quotient of a space form (i.e., a sphere, Euclidean space, or a hyperbolic space). Wang showed that for $n \neq 4$ the largest dimension of the isometry group of an $n$-manifold that is not a quotient of a space form (i.e., isomorphic to a sphere, a Euclidean plane, or hyperbolic space) is exactly $${{n - 1} \choose 2} + 1 .$$
The case $\dim n = 4$ is an exception, as we have $\dim \operatorname{Iso}(\Bbb C \Bbb P^2) = \dim \operatorname{PSU}(3) = 8$. But $\Bbb C \Bbb P^2$ is not spin, so it doesn't even embed smoothly (let alone isometrically) into $\Bbb R^6$.
Remark In the special case $p = n - p = 1$, our construction yields the usual embedding of the flat torus $S^1 \times S^1$ into $\Bbb R^4$.