Partitioning a rectangle into different isosceles triangles
As Noam Elkies has observed, any acute non-isosceles triangle can be tiled by three pairwise non-congruent isosceles triangles, by connecting each vertex to the circumcenter. There are lots of ways to partition any rectangle into non-congruent non-isosceles triangles, each of which can be replaced by three isosceles triangles, and I think it should be easy to find a partition for which this construction produces non-congruent isosceles triangles.
Let $\ A\ B\ C\ D\ \in\ \mathbb R^2\ $ be the vertices of a rectangle, where $\ A+C=B+D=\mathbb 0\ $ is the origin. Let $\ E\ $ belong to the interval $\ BD,\ $ and be such that $\ AE\ $ and $\ BD\ $ are perpendicular one to another.
Then we get the following partition of the square into six isosceles triangles (the vertices on the symmetric line are listed as the middle of the three vertices):
$$ A\ \ \frac{A+B}2\ \ E $$ $$ B\ \ \frac{A+B}2\ \ E $$ $$ A\ \ \frac{A+D}2\ \ E $$ $$ D\ \ \frac{A+D}2\ \ E $$ $$ B\quad \mathbb 0\quad C $$ $$ C\quad \mathbb 0\quad D $$
Thus the problem is solved, with $\ \mathbf 6\ $ triangles, in the case of all rectangles but squares--only in the case of a square some of the given $\ \mathbf 6\ $ triangles are congruent. Otherwise, we get three pairs of triangles which have the same area within the pair but different for the different pairs. And within the pair, one triangle is acute (i.e. all its angles are acute), and one is obtuse. Thus no two of the six are congruent.
In the case of square, @Wolfgang's construction provides $7$ triangles. However, $\ \mathbf 5\ $ is enough.
Indeed, let
$$ a\ :=\ 2-\sqrt{2}\ =\ \sqrt{2}\cdot(\sqrt{2}-1) $$
Then, consider the following isosceles triangle decomposition of square $[0;1]^2$:
$$ (0\ 0)\quad (0\ 1)\quad (1\ 1) $$ $$ (0\ 0)\quad (a\ 0)\quad (a\ a) $$ $$ (a\ 0)\quad (a\ a)\quad (1\ 1) $$ $$ (a\ 0)\quad (\frac{a+1}2\,\ \frac 12)\quad (1\ 0) $$ $$ (1\ 0)\quad (\frac{a+1}2\,\ \frac 12)\quad (1\ 1) $$
Not any two of them are congruent.