I unroll the tetrahedron. Intuitively, I prefer vertices, symmetry and equal distances. I use the theorem of Pythagoras.

2 pirates
Place one pirate in the center of a face, and the other one in the opposite vertex. The distance between these pirates is 1.1547005...

3,4 pirates
Place the pirates in 3 (4) vertices. The distance between any two pirates is 1.

5 pirates
Place one pirate in the center of face BCD, and one in A. Place three pirates on the sides AB, AC and AD, at a distance 1/3 from B,C,D apart, respectively. The minimal distance between two pirates is 2/3.

6 pirates
Place two pirates near the centers of faces ABC and BCD, but a little bit further away from side BC. Place two pirates in B and C. Place two pirates near A and B, but a bit closer to each other.
The minimal distance between two pirates is 0.5898647...

7,8 pirates
Place pirates in all vertices and three (four) centers of faces. The minimal distance between two pirates is 0.577350....

9,10 pirates
Place pirates in all vertices and the middle of five (six) sides. The minimal distance between two pirates is 1/2.

Jack Burns