Clifford torus as the fixed point set of an isometry from $S^3\rightarrow S^3$? 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 $\phi$?
 A: 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, 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.
