Cerium is the oxidant or not in this reaction (probably is), but cerium oxalate should be Ce2(C2O4)3 in the former case or Ce(C2O4)2 in the later one.
These reaction can be balanced even without procedure for redox reaction taking into account hydrogen atoms on the left side and water on the right side.
Follow up my way of thinking!
You need 2 cerium atoms on the right side. Put 2 on the left one.
2Ce(IO3)4 + H2C2O4 –> Ce2(C2O4)3 + I2 + H2O + CO2
you have 2 cerium atoms, 8 iodine atoms and 24 oxygen atoms that need 48 hydrogen atoms to form water, hence you need 24 molecules of oxalic acid, 3 of them form cerium(III) oxalate, the rest molecules of oxalic acid forms carbon dioxide. So the result is
2Ce(IO3)4 + 24H2C2O4 –> Ce2(C2O4)3 + 4I2 + 24H2O + 42CO2
In the second case, thinking analogously:
Ce(IO3)4 + 12H2C2O4 –> Ce(C2O4)2 + 2I2 + 12H2O + 20CO2