This is I guess a common problem that stems from the way we use the word "intensity". Normally, we would use intensity to mean something like the energy going through a unit area per unit time. In the case of the photoelectric effect, we instead generally mean the number of photons (per unit time), as it is this that decides how many photo-electrons are emitted, and directly determines the current.
Also note that there is no way we can have a "small" change in the number of photons, compared to unity. If you slightly increase the frequency of a monochromatic source (a phrase that typically means a source of fixed frequency) by an arbitrarily small amount, the energy-intensity will necessarily increase. Only when you get to the point where the increase in frequency can be compensated for by reducing the number of photons by exactly one is it possible to have the same energy-intensity as what you started out with. You can see that energy-intensity is somewhat more complicated to keep constant when you bring in photons.
Anyway, for such cases, all you have to do is look at the variation of saturation current with frequency for a fixed photon-number-intensity i.e. consider the frequency and photon number as independent variables.
EDIT: For the saturation current, we consider the case when all the electrons get there, and the number of these electrons (per unit time), which is the current, is given by the photon number. Making the electrons get there is the precise function of the anode voltage in these situations, and the saturation current appears with a very large voltage. Otherwise, the kinetic energies of the electrons is the basic reason behind the variation of current with the anode voltage.