I apologize in advance for the wall of text.
This solution will use the full length of the rope, assuming that one twelfth of the rope is within the distance of one individual's arm's reach. If the rope is longer, the remainder can be ignored, as only one end-point is needed and not using the entire rope is within the requirements of the puzzle.
For sake of notation, one end of the rope will be referred to as point A and the other end will be referred to as point B. The section of rope between two points will be referred to as the point names, e.g. section AB is the section of rope from point A to point B. A single unit will be one twelfth of the length of section AB. People will be referred to as P1, P2, and P3. For diagrams, each dash (-) represents one one unit of rope and the holders of the point will be marked.
1: measure one third of section AB. This can be done by having P1 tightly hold point A and loosely hold a bend in the rope close to point B, while P2 is tightly holding point B and loosely holding a bend in the rope close to point A. As they walk away from each other and pull the rope taut, the loosely held sections will slip through their hands to separate the rope into 3 parts. The loosely held points will be at 4 units and 8 units from point A. The point at 4 units from A will now be known as point C.
2: P1 will drop their loosely held point. P2 will give point C to P1's free hand. P3 will loosely hold section AC and stretch the rope taut to discover the center point of section AC. This midpoint will be D.
3: C will be given to P3's free hand and D will be given to P1's now free hand. Using his newly freed hand, P3 will repeat the above process to find the midpoint of section AD, which will be point E. P1 will drop D and take E from P3.
4: P1 and P3 will pull the section CE taut. P2 will run section CB along the section CE and hold the point at which it meets point E, which will be point F. After F is found, P1 will drop E.
5: At the point, section AB is separated into 3 sections, of lengths 4, 3 and 5 units. P2 will give P1 F and afterwards, receive A from P1.
6: P2 will hold points A and B together and all three sections will be pulled taut. The only way for all three section to be taut is for the angle at point C be a right angle. At this point, the rope should be on the ground to prevent it from moving while not attended. P2 will abandon A in its current position.
7: P2 will run section FB along section FC. Once P2 reaches C, P3 will hold the point where FB reaches C. This will be point G.
Note: C and G are at the same location
8: P2 will run section GB along section CF, holding the point along CF where B reaches. This will be point H. P3 can now drop point G and P1 will hold point point H, while making sure not to move point F (this is the reason for the one unit being within arm's reach). P2 will move back to A and hold onto it. Given that C and F have not moved, the right angle can be confirmed again.
9: This is where it gets tricky. P2 will loosely hold a point at least one unit from point A along section AC to ensure that the angle remains unchanged. P3 will ensure that the rope does not leave point C, but will allow it to slide freely through that point. P1 will keep their hand on the ground where point F is. After this preparation, P1 will pull point H to where their hand marked point F on the floor and hold it there. This will cause AC to shorten by 1 unit (the length of section HF). P2 will adjust B to the new position of A.
10: P1 will use their free hand to swing the loose section HB in P3's direction. P3 will catch it, while making sure not to move C, and pull HB against point C, marking the point where section HB meets point C. This will be point I.
11: P3 will leave C and pull point I to the opposite corner until it becomes taut. This should result in a square.