Given that in $\triangle ABC$, $A$, $B$, and $C$ are internal angles, and $\sin A \cos A = \sin B \cos B$,

Using the double angle identity for sine, we can rewrite the equation as $\frac{1}{2} \sin 2A = \frac{1}{2} \sin 2B$, which simplifies to $\sin 2A = \sin 2B$.

This implies that either $2A + 2B = 180^{\circ}$ or $2A = 2B$.

Simplifying these equations, we get $A + B = 90^{\circ}$ or $A = B$.

Therefore, $\triangle ABC$ is either an isosceles triangle (when $A = B$) or a right triangle (when $A + B = 90^{\circ}$).

So, the answer is: $\boxed{\text{Isosceles or right triangle}}$.

The given equation was simplified using the double angle identity for sine to determine the relationship between $A$ and $B$, which allowed for the conclusion to be drawn. This problem tests understanding of the double angle identity for sine, so being proficient with this identity is key to solving this problem.