Five equal forces equally inclined to neighboring forces act on a point. Find the angle between any two. Or its spherical co-ordinate $ \theta, \phi $ decomposition.
For $n=4,6,$ the angles are $ -\cos^{-1}\frac13, \pi/2 $ respectively ( tetrahedron and cube). But $ n=5$ is not among the Platonics.
I have not done an experiment but am reasonably convinced that five such forces acting on a knot and weighed down by five equal forces through suitably arranged smooth pulleys can result in static equilibrium.
Because you do not have a platonic solid with exactly five vertices, you can't have five force vectors that are all interchangeable under point-group symmetry and the resultant being zero unless you drop down to two dimensions making a regular pentagon.
To get the vectors "as symmetric as possible" we might try a regular pentagon, a square pyramid or a triangular bipyamid. All angles shown are for cases where all vectors are equal in magnitude and give a zero resultant.
Regular pentagon: five of the ten angles between pairs of vectors measure $72°$, the other five measure $144°$.
Square pyramid: four of the angles measure $86.4°$, four measure $104.5°$, and two measure $151.0°$. Rounded to nearest tenth.
Triangular bipyramid: six angles measure $90°$, three measure $120°$, and one measures $180°$.
Which is most "nearly equiangular"? The pentagon uses only two distinct angles, but no more than five pairs of vectors are equiangular to each other and cramming all five vectors into one great circle while the rest of the sphere is "empty" looks ugly. The triangular bipyramid has the six smallest angles all equal and nine out of ten within $30°$ of each other, but that one straight angle sticks out like a sore thumb. The square pyramid has the smallest difference between minimum and maximum angles but only four angles equal to each other; the nearest neighbors along the base are not equivalent to the neighbors between axial and basal vectors. In the end it's really a judgment call.
My vote is for the triangular bipyramid.