What is the product of the titration of potassium permanganate and iron(II) sulfate?
From your description I'd say you were titrating ferrous sulphate, $\ce{FeSO4}$ solution (the analyte), with potassium permanganate, $\ce{KMnO4}$ solution (the titrant), in acid conditions (dilute $\ce{H2SO4}$ present).

The permanganate ion $\ce{MnO4^{-}}$ is a strong oxidiser and oxidises the ferrous ion $\ce{Fe^{2+}}$ to the ferric ion $\ce{Fe^{3+}}$ very easily. In dilute solution both ferrous and ferric ions are more or less colourless.

In this process the very strongly coloured permanganate is reduced to the manganous $\ce{Mn^{2+}}$ ion, which is also colourless in dilute solution.

When all the ferrous ions have been oxidised, one more drop of the strongly coloured permanganate solution then turns the analyte solution pink and this is your end point.


  I recently did an experiment at school, where I had to titrate $\ce{KMnO4}$
  with $\ce{FeSO4}$.


In that case you'd have started from a purple solution (analyte) which would gradually, during the titration, have turned colourless. This method is also used.



As this is a redox reaction, I suggest you write out (or look up) the reduction (permanganate to manganous) half reaction and the oxidation (ferrous to ferric) half reaction, to complete your knowledge of this titration.