Let $S^3=\{(z,w)\in {\mathbb{C}}^2:|z|^2+|w|^2=1\}$, 
and $T\_{\pi/4}^2:=\{(e^{i\alpha}/\sqrt{2},e^{i\beta}/\sqrt{2}):\alpha,\beta\in \mathbb{R}\}$. 
Is there an isometry $\phi:S^3\rightarrow S^3$ whose fixed point set (i.e., 
$\{p\in S^3:\phi(p)=p\}$) is exactly $T\_{\pi/4}^2$? If yes, then what is an explicit expression for $\phi$?



Clifford torus as the fixed point set of an isometry from $S^3\rightarrow S^3$?
An isometry of the sphere with respect to the restriction of the Euclidean distance metric (turns out to be) the same as an isometry of the sphere with respect to the [great-circle metric](https://en.wikipedia.org/wiki/Great-circle_distance), which is the induced distance of the round Riemannian metric on the sphere (restriction of Euclidean Riemannian metric). The great-circle distance between $x$ and $y$ in $S^n \subset \Bbb R^{n+1}$ is precisely $\text{arccos} \langle x, y \rangle$.


If $\phi$ is an isometry of $S^n$, then there is an extension $\tilde \phi: \Bbb R^{n+1} \to \Bbb R^{n+1}$, with $\tilde \phi(tx) = t\tilde \phi(x)$ for $x \in S^n, t \in \Bbb R\_{\geq 0}$. 


Now by assumption we have $$\langle\tilde \phi(tx), \tilde \phi(sy)\rangle = ts \langle \phi(x), \phi(y) \rangle = ts \langle x, y\rangle = \langle tx, sy\rangle.$$ 


A mapping which preserves inner products clearly preserves norms, but more importantly preserves norms of *sums* by the polarization identity $$\|x+y\|^2 = \|x\|^2 + \|y\|^2 + 2\langle x,y\rangle.$$ Preserving norms of sums (equivalently, differences) means that $\tilde \phi$ is an isometry, and isometries of Euclidean space that fix 0 are linear. 


In particular, the fixed points of $\phi$ are the fixed points of $\tilde \phi$ on the sphere $S^n$. The fixed points of $\tilde \phi$ form a linear subspace; intersecting that with the unit sphere gives a linear sub-sphere of $S^n$. The only fixed point spaces of isometries of $S^n$ are unknotted spheres.


Ryan Budney's comment explains why the Clifford torus is not even the fixed point set of any finite order diffeomorphism: it would need to swap sides of the torus, but given that loops in the torus (which would be fixed under such a diffeomorphism) may bound on one side but not the other, this is impossible. This implies that the Clifford torus is not the fixed point set of an isometry with respect to *any* metric on $S^3$. 


Actually, finite order diffeomorphisms of $S^3$ are *conjugate* to a linear isometry by a form of elliptization for orbifolds, so the same result above applies to the possible fixed point sets. This fails already for $S^4$, which has many involutions with fixed-point-sets various 3-manifolds $\Sigma$ with the same homology as $S^3$ but complicated fundamental group.


