Let $T$ be the triangle in the coordinate plane with vertices $\left(0,0\right)$, $\left(4,0\right)$, and $\left(0,3\right)$. Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}$, $180^{\circ}$, and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)
First, it is critical to note that the number of reflection transformations must be even in order for the triangle to return to original position. This is because a reflection will change this triangle's orientation. Hence, let's discuss two different scenarios: no reflection and two reflections.
If there is no reflection, then it must have three rotations. In order to make the triangle return to the original positions, the total rotated angle must be a multiple of $360^{\circ}$. Therefore, let's investigate the non-negative integer solutions to the following indeterminate equation: $$90 x + 180 y + 270 z = 360 k$$
where $k = 1, 2$ and $x + y + z = 3$. Replacing $x$ and simplifying the above relation will yields $$y + 2z = 1\quad\text{or}\quad 5$$
There are only a limited number of possibilities. Trying all of them will find the following solutions: $$(x, y, z) = (2, 1, 0), (0, 1, 2)$$
It follows that the number of combinations to make the triangle return to its original position is all the permutations of these rotations: $2 \times 90^{\circ} + 1\times 180^{\circ}$ or $1\times 180^{\circ} + 2\times 270^{\circ}$. There are totally $\underline{6}$ cases because each has $3$ possibilities by using the multiplication principle (there are three choices for the rotation which appears once).
If there are two reflections, then these two must be different. Otherwise, applying the same reflection twice will turn the triangle to its original position which makes it impossible to remain the same place after a third transformation. Applying two different reflections, regardless of the sequence, is equivalent to rotating a $270^{\circ}$. The only way to make the triangle return to its original position is to make another $270^{\circ}$ rotation. There are $3\times 2\times 1= \underline{6}$ ways to arrange the sequence of these three distinct transformations.
Hence, all in all, there are totally $6+6=\boxed{12}$ different combinations of transformations satisfying the requirement.