A partial solution
You asked for the existence of a $(2n+1)$-sided polytope/die, so that all $2n+1$ sides are congruent and so that the probability of landing on each of these sides is precisely $1/(2n+1)$.
I do not see how to mathematically model the property that the probability of landing on each of the sides is precisely $1/(2n+1)$.
This seems to need some assumptions from physics and/or mechanics.
Instead, I propose a purely mathematical formulation of fairness:
Definition of fairness: A polytope is a fair die, if for any two sides there exists a symmetry of the polytope that maps one side into the other one.
This fairness definition works for the standard six-sided die (which is highly symmetric) and also for the 10-sided polytope that you describe in your puzzle.
Note that the definition trivially implies congruence of any two faces.
I will prove below that for this definition of fairness, there does not exist a fair die with an odd number of sides.
Non-existence proof for an odd number of sides
(1) Suppose for the sake of contradiction that there exists a $(2n+1)$-sided polytope that is fair.
Let $s$ denote the number of edges of every side.
The polytope has altogether $2n+1$ sides, and every side has $s$ edges.
Hence the total number of edges is $(2n+1)s/2$, as every edge is counted twice.
The polytope has $e=(2n+1)s/2$ edges and $s\ge3$ is an even integer.
(2) Consider a side of the polytope.
The side has $s$ vertices, and we let $d_1\le d_2\le\cdots\le d_s$ denote the number of edges that are incident to these $s$ vertices.
A vertex with $d_i$ incident edges is a vertex of $d_i$ different sides.
Hence, the polytope contains exactly $(2n+1)/d_i$ vertices of this particular type.
We conclude:
The polytope has $v=(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s})$ vertices.
(3) Next we use Euler's polyhedral formula that says that a polytope with $f$ sides (faces), $e$ edges, and $v$ vertices must satisfy $v+f-e=2$.
By plugging in $f=2n+1$ and the expressions derived in (1) and (2), we get
$(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s}+1-\frac{s}{2})=2 ~~~~~~(*)$
Since $d_i\ge3$ for all $i$, we furthermore derive
$(2n+1)(s\frac{1}{3}+1-\frac{s}{2}) ~\ge~ (2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_s}+1-\frac{s}{2}) ~=~ 2 $
Since $2n+1$ is positive, also the value $1-s/6$ in the other bracket must be positive; this implies $s\le5$.
Since $s\ge3$ is an even integer, we arrive at the following fact:
$s=4$
(4) Since $s=4$, the equation $(*)$ now simplifies to
$(2n+1)(\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+\frac{1}{d_4}-1)=2 ~~~~~~(**)$
If $d_1\ge4$, then the sum of the four reciprocals in $(**)$ would be at most $1$; a contradiction. We conclude $d_1=3$, and $(**)$ further simplifies to
$\frac{1}{d_2}+\frac{1}{d_3}+\frac{1}{d_4} ~=~ \frac{2}{3}+\frac{2}{2n+1} ~~~~~~(***)$
This only leaves four possible cases for $d_2$ and $d_3$
(in all other cases, the sum of the three reciprocals would be at most $2/3$):
- (a) $d_2=3$ and $d_3=3$
- (b) $d_2=3$ and $d_3=4$
- (c) $d_2=3$ and $d_3=5$
- (d) $d_2=4$ and $d_3=4$
(5) It remains to do the case work.
In case (a), equation $(***)$ turns into $d_4=(2n+1)/2$; this is impossible, as $d_4$ would not be integer.
In case (b), equation $(***)$ turns into $d_4=12(2n+1)/(2n+25)$.
Then the odd number $2n+25$ must divide $3(2n+1)=6n+3$.
Since $6n+75=3(2n+25)$, also the difference $6n+75-(6n+3)=72$ must be a multiple of the odd number $2n+25$; a contradiction.
In case (c), equation $(***)$ turns into $d_4=15(2n+1)/(4n+32)$.
Then the odd number $15(2n+1)$ must be a multiple of the even number $4n+32$; another contradiction.
In case (d), equation $(***)$ turns into $d_4=6(2n+1)/(2n+13)$.
Then the odd number $2n+13$ must divide $3(2n+1)=6n+3$.
Since $6n+39=3(2n+13)$, also the difference $6n+39-(6n+3)=36$ must be a multiple of the odd number $2n+13$; the final contradiction.
(6) As all possible cases have ended up in a contradiction, we conclude that there is no fair die with an odd number of sides.
