How to convince myself (imagine) that $\Bbb S^1$-action on $\Bbb S^3$ fixes a circle of sphere?

One would not expect a rotation in $\mathbb C^2 \approx \mathbb R^4$ to have an "axis of rotation" which is a line, i.e. something of real dimension $1$. On the other hand, one would expect the "axis of rotation" to have real codimension $2$, which it does: the entire plane $w_1=0$ is fixed. And when you intersect that plane with $S^3$ you get a circle that is fixed.

If you want to visualize this example, it can be done using the fact that $S^3$ is the one-point compactification of $\mathbb R^3$, which I'll write as $S^3 = \mathbb R^3 \cup \{\infty\}$. In this model, one can visualize the circle of fixed points as the unit circle in the $x,y$-plane: $$\{(x,y,0) \mid x^2 + y^2 = 1\} $$ Outside of this circle of fixed points, every other orbit of the action is a circle, and one can visualize these circle orbits in $\mathbb R^3 \cup \{\infty\}$ using $(r,\theta,z)$ cylindrical coordinates, as follows. One of the circle orbits is $\text{$z$-axis} \cup \{\infty\}$. Then, for each constant angle $\theta_0$, the half plane $\theta = \theta_0$ pierces the fixed circle at the single point $P(\theta_0)$ with coordinates $(r,\theta,z)=(1,\theta_0,0)$, the boundary edge of that half plane is the $z$-axis which is an orbit, and the rest of the half-plane is foliated by a family of circle orbits which approach that single point in one direction getting smaller and smaller, and which approach the $z$-axis in the other direction getting larger and larger (in the hyperbolic metric $\frac{dr^2+dz^2}{r^2}$ on this half-plane, these are the concentric circles centered on $P(\theta_0)$).