We are given the digits $\{0,2,4,6,8\}$ and no digit may repeat.

We need the number of all numbers greater than $10{,}000$.

Since the available digits are only 5 digits total, any number greater than $10{,}000$ formed from these digits without repetition must be a 5-digit number.

So we count all 5-digit numbers that can be formed using all five digits $0,2,4,6,8$ exactly once, with the condition that the first digit cannot be $0$.

Total permutations of the 5 digits: $5! = 120$

Count the invalid arrangements where $0$ is in the first place. Fix $0$ in the first position, then arrange the remaining 4 digits: $4! = 24$

Therefore, valid 5-digit numbers are $120 - 24 = 96$

Hence, the number of all numbers greater than $10{,}000$ is $\boxed{96}$