It is possible to find examples of two or more non-congruent right triangles with the same hypotenuse. I don't know of a systematic way to search for them (I used brute force). A few thoughts directed me to a choice of a hypotenuse to research: the hypotenuse of a primitive right triangle with integer sides must be odd; and that odd number may not have an odd number of prime factors congruent to $3\mod(4)$ if it can be represented as $m^2+n^2$. I chose to look at $1105=5\cdot 13\cdot 17$, which yielded the following three primitive Pythagorean triples: $(1105,1104,47),\ (1105,1073,264),\ (1105,943,576)$.
In the $m^2-n^2,\ 2mn,\ m^2+n^2$ representation of such triples, the first corresponds to $m=24,\ n=23$; the second to $m=33,\ n=4$; and the third to $m=32,\ n=9$.
Added by edit: Fermat proved that every prime number of the form $4n+1$ can be expressed as the sum of two squares in only one way. Primes of the form $4n+3$ cannot be expressed as the sum of two squares.
The Brahmagupta–Fibonacci identity says that $$(a^2 + b^2) (c^2 + d^2) = (a c − b d)^2 + (a d + b c)^2 = (a c + b d)^2 + (a d − b c)^2$$
This means that two suitable primes, each of which can be represented uniquely as the sum of two squares, have a product that can be represented in two distinct ways as the sum of two squares. Thus any odd number that is the product of two distinct primes, each of which has the form $4n+1$, can be the hypotenuse of two distinct primitive Pythagorean triples (as exemplified in the comment by David K). The inclusion of more suitable prime factors in the number corresponding to the hypotenuse will increase the number of ways that the resulting product can be represented as the sum of two squares (as exemplified in my earlier given example).