i remember visiting this question a while back and there was a long discussion going on... why was it deleted? anyways i tried to come up with a solution to this and would like to see some other approaches.. create an axis of rotational symmetry and assign a top and bottom to the axis where it intersects the sphere. draw a line from top to bottom s.t the line never reverses direction, except perhaps horizontally (*for short distance) create four more copys of this line and rotate them about the axis of symmetry and space them evenly s.t 5 congruent regions are formed, each line belongs to one of the regions (*sometimes only partially). now for a more flexible creation we can consider the surface of the sphere to be composed of infinitesimally thin rings (slice the sphere horizontally from each point of the axis of symmetry) divide each slice/ring into 5 segments of equal length to form 5 pieces of our congruent regions, one of each. instead of using lines to form 5 congruent regions we can take each ring and color the 5 segments a different color (always in same order) and rotate each about the axis s.t each ring touches the ring below and above with at least one point of the same color. this makes sure the regions are connected. now suppose we have any 5 points on the sphere, we place an axis through it s.t no 2 of the 5 points are on the same horizontal line (same ring) this is easy to do. call the points $ A,B,C,D,E$ and the regions $ 1,2,3,4,5$. attach a different number to each point to tell us the region its in eg: $ A1,B2...$ to each point there corresponds a point in each different region (in the same ring) so we have the points $ A1,...,A5,B1,...,B5,...$ as long as the points with the same letter follow the same order of the numbers and direction (clockwise/anti) around their respective rings then we can create 5 congruent regions that satisfy our goal. all we have to do is insert the intermediate colored rings, and we can easily connect two points which span a distance of half the sphere beacuse the rings are infinitesimally thin. to show that we can always find an orientation s.t no 2 points are on the same horizontal line (on a surface) we can take $ n$ points and draw all lines connecting any 2 of them. take a point on the surface now and draw all lines through it that are perpendicular to a line we drew between two points. there are only a finite amount of these lines, so we can chose a line through the chosen point s.t no 2 points from our set are on the same horizontal. one thing that bothers me is what about the points at the top and bottom of the axis of symmetry of the sphere, they do not have 4 corresponding points each...? and i was also wondering if you can split the surface of the sphere into congruent regions that do not have an axis of symmetry? i think not, but how can this be proved.

I think not always.Assign axis for Mars like the earth's axis.Also define the north and south pole on Mars.Draw a sufficiently small circle centered at the north pole,put the research lab one by one an half of the circle.And any $2$ of the labs are sufficiently close to each other.Suppose there are $5$ connected congruent region whose union covers the Mars and each one of them contain exactly $1$ lab.Look at the front elevation,from left to right,denote the $5$ labs be $A,B,C,D,E$,draw $2$ vertical lines $l_1,l_2$ each from $A,E$.Call the area between the $2$ lines a strip,call the region covers $A,B,C,D,E$ be $S_1,……,S_5$.We know that $S_1,……,S_5$ pairwise disjoint,so if $S_2$ bent out of the strip at some point on $l_1$ called $P$,either it brokes into parts or not congruent with $S_1$ or have common point with $S_1$,since $A,B$ are not concided on the circle around the pole.So $S_2,S_3,S_4$ all in strip,but $S_1,S_5$ must have portion out of strip,they cannot be congruent. Maybe somewhere wrong,oooop