I mean 3-sphere (normal, like Earth) has 3 euators: namely equator, 0h meridian circle and 6h meridian circle.
So, "pole" is a point, where all coordinates equal zero, except one, which equals to sphere radius.
Can't factor out, how many such equators 4-sphere has?
On 3-sphere each equator intersects with 2 other equators in 4 poles. In each pole 2 equators intersect.
On 4-sphere there should be 3 equators intersecting in a pole. These 3 equators should also intersect at opposite pole.
So we have
$E_1=\{P_1, \bar{P_1},...\}$
$E_2=\{P_1, \bar{P_1},...\}$
$E_3=\{P_1, \bar{P_1},...\}$
where equator $E_i$ is represented with a set of poles it contains, while pole is denoted by $P_j$, having $\bar{P_j}$ as opposite pole.
There should be at least one more equator, which intersects with three previous:
$E_4=\{P_2, \bar{P_2}, P_3, \bar{P_3}, P_4, \bar{P_4},...\}$
poles $P_2...P_4$ should be on previous equators, so we have
$E_1=\{P_1, \bar{P_1}, P_2, \bar{P_2}, ...\}$
$E_2=\{P_1, \bar{P_1}, P_3, \bar{P_3}, ...\}$
$E_3=\{P_1, \bar{P_1}, P_4, \bar{P_4},...\}$
What we should have at ellipsis? Seems that it should be
$E_1=\{P_1, \bar{P_1}, P_2, \bar{P_2}, P_3, \bar{P_3}\}$
$E_2=\{P_1, \bar{P_1}, P_3, \bar{P_3}, P_4, \bar{P_4}\}$
$E_3=\{P_1, \bar{P_1}, P_4, \bar{P_4}, P_2, \bar{P_2}\}$
but I can't imagine, how 2 equators can intersect 4 times???