Is this the only solution for $|r_k-r_{k'}|=\text{const}, k \neq k'$ for $k=0,..,4$ for 3 dimensions? Is the only solution for 5 vertices with relations
$|r_k-r_{k'}|=\text{const}$ for $k \neq k'$, where $k=0,1,2,3,4$ for 3 dimensions?
Here I am writing in bracket notation i.e. for unit vector $x_i$, the vector $r_k=ax_0+bx_1+cx_2=\langle a,b,c\rangle$
$r_0=\langle 0,1,0\rangle$
$r_1=\langle 1,0,0\rangle$
$r_2=\langle 0,-1,0\rangle$
$r_3=\langle -1,0,0\rangle$
$r_4=\langle 0,0,1\rangle$
If so why?
Edit: this arrangement doesn't satisfy the condition above, but does leave the possibility answer open
 A: Could you clarify the question please?
What norm are you using: euclidean norm? Is $|r_0-r_1| = |(-1,1,0)| = \sqrt{2}?$
 But then $|r_1-r_3| = |(2,0,0)| = 2$.   I do not know if I can really follow; and why wouldn't $r_5= (0,0,-1)$ be a possibility?
If I understood the question, then you ask for $S\subset \mathbb{R}^3$, such that for each two different $r_k,r_{k'} \in S$ it holds true that $|r_k-r_{k'}| = const$.
Before you continue reading, it all summarizes to:
I think there are only $4$ such vertices. And they form a regular tetrahedron, because all the diagonals of the polygon need to be as long as its sides. Then there cannot be $5$ as that would imply the height of the Tetrahedron to be half it's side length, which is not the case. Now I left the rest here, because it made fun thinking about the problem in that way.
Now let $C:=\{(r_k-r_{k'}) \mid r_k,r_{k'} \in S, k \neq k'\} \subset \mathbb{R}^3$. Then, $| \cdot |$ has to be constant on $C$, so $C$ has to be a ball around $0$.
Lets start by considering only three points in $S$. Fix $r_1,r_2,r_3\in S$, and let $C_i:= C + r_i$ be the ball shifted by $r_i$, $i \in \{1,2,3\}$. Then it must hold true that:
$r_i,r_j \in C_k \text{, where } i,j,k \in \{1,2,3\}$  pairwise distinct. This means that the $3$ balls are pairwise intersecting, and $C_i$ passes through the midpoints of $C_j,C_k$, which are $r_j,r_k$. So $C_1,C_2,C_3$ are pairwise intersecting and each contain the midpoint of the others. 
So it boils down to the question: How many equally large balls can we place in $3$-space that touch pairwise and all contain the midpoint of the other?
And well: two balls ($C_1,C_2$) cut in a circle $C_1 \cap C_2$, so the third midpoint $r_3$ must lie on that cutting circle. Now by construction, $r_1,r_2,r_3$ are equally distant. Look at $C_3$, it cuts the circle given by $C_1 \cap C_2$ again in $2$ points $u$ and $v$. This leaves us with 2 possible points to choose the midpoint $r_4$ of the fourth ball, as otherwise they are not equally distant and don't pairwise contain their midpoints. Wlog let $r_4 = u$. Now by construction the fourth ball $C_4$ goes through the other $3$ midpoints, and they form a regular tetrahedron. The question remains why $C_4$ does not pass through $v$, since if it would one could chose $r_5=v$ and would have $5$ equally distant points. Maybe someone can clarify here? (Note that this proves already that there cannot be more than $5$ points as there are no intersection-points left.) I would have to think a bit longer for a more rigorous proof here. 
