Why does the nature and oxidation state of a central metal affect field split?
Let me start this answer — as usually for an answer I give about coordination complexes — by showing the molecular orbital scheme of a typical octahedral complex ignoring $\ce{L\bond{-&gt;}M}$ &#x3c0; interactions.


Figure 1: Octahedral $\ce{[ML6]}$ complex with no &#x3c0; interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

It is easy to see where field splitting comes from: two of the ligand group orbitals transform as $\mathrm{e_g}$ as do two of the metal’s $\mathrm{d}$ orbitals ($\mathrm{d}_{x^2-y^2}$ and $\mathrm{d}_{z^2}$). These interact in a bonding/antibonding manner to give $\mathrm{e_g}$ (considered a ligand-centred orbital due to the electronegativity difference) and $\mathrm{e_g^*}$ (a metal-centred orbital).

It is important to note that this scheme depends strongly on the relative positions of the orbitals. The closer the energies of the ligand and metal group orbitals, the larget the split will be.

If you consider different aquacomplexes, you can always be sure that the ligand orbitals start off from the same energy (they are the same water, after all). However, the metal orbitals will be differently positioned. The higher the first ionisation enthalpy (typically corresponding to the removal of an electron from $\mathrm{4s}$ — here $\mathrm{a_{1g}}$) the lower these orbitals will be in this scheme. Loosely, the ionisation enthalpy depends on the number of protons in the nucleus, so we might expect nickel to have lower-lying orbitals than titanium. Whatever the actual difference is (there are many more effects), no two different metals will ever have the same orbital energies and therefore any two different metals will always have different field splits corresponding to a different colour. You know this: hexaaquachromium(III) is green while hexaaquairon(III) is yellow-orange.

The oxidation state influences the orbital scheme in a similar way. Each oxidation corresponds to the removal of an electron meaning less electronic repulsion meaning lower-lying orbitals (in slightly larger contexts, this corresponds to the contraction of cations). Thus, if you change the oxidation state of a metal, e.g. from $\mathrm{+II}$ to $\mathrm{+III}$, the metal orbitals will be lowered in energy and the field split will most likely increase since there is better overlap with the ligand orbitals. Therefore, iron(II) and iron(III) have different colours in solution.