Can a continuous map $S^2 \rightarrow S^2$ preserve orthogonality without being an isometry?

Suppose I have a map $\phi: S^2 \rightarrow S^2$ and I know that

a) $\phi$ is continuous and bijective

b) If $a$ and $b$ subtend an angle of $\pi / 2$ at the center of the sphere, then so do $\phi(a)$ and $\phi(b)$.

Does it follow that $\phi$ is an isometry of the sphere?

If yes, can you sketch a proof?

If no, can you furnish an example of a non-isometry $\phi$ satisfying the above?

(Also, in either case, is the requirement that $\phi$ be bijective redundant?)

In the case that you are willing to assume that the map is once differentiable, the answer is yes.

Sketch of proof:

There is a duality between great circles on $\mathbb{S}^2$ and lines in $\mathbb{R}^3$: you can identify a great circle on $\mathbb{S}^2$ with plane in $\mathbb{R}^3$ that contains the great circle, and then identify with the line that is normal to the plane. In other words, you can write $C_v = \{ w\in \mathbb{S}^2 | w\cdot v = 0 \}$ for any $v\in\mathbb{S}^2$.

The preservation of orthogonality means, therefore, that your map $\phi$ sends great circles to great circles.

Furthermore, we observe that fixing a point $p\in \mathbb{S}^2$, we can identify its tangent directions with the collection of all great circles through it. For $\eta,\omega\in T_pM$, the angle between them can be measured by the angle between their corresponding great circles, which is the same as the angle between their corresponding dual vectors.

So: if $\phi$ is $C^1$, the differential $d\phi$ defines a linear map between the tangent spaces. That $\phi$ preserves orthogonal directions now implies that $d\phi$ preserves orthogonal directions. Hence $d\phi$ must be conformal! (Since it is linear and preserves orthogonality.) So $\phi$ is a conformal map of the sphere. But recall back that $\phi$ preserves all great circles--any conformal automorphism of the sphere that preserves all great circles must be an isometry.

Actually, the answer is positive even if you assume only that $\phi$ is a bijection (no need to assume continuity). The point is that such $\phi$ sends great circles to great circles and antipodal pairs to antipodal pairs. Hence, $\phi$ projects to a bijection $\Phi: RP^2\to RP^2$ which preserves collinearity. Von Staudt proved (the fundamental theorem of projective geometry) that all such maps $\Phi$ are projectivizations of linear transformations $R^3\to R^3$. You can find a proof for instance, in Hartshorne's book "Projective Geometry". Now, it is an elementary exercise to prove that a projective transformation which preserves distances equal to $\pi$ on $RP^2$ has to be an element of $PO(3)$. Hence, $\phi$ is orthogonal.