In CuSO4. 5H2O, water molecules acts as ligands providing free electrons to Cu2+ ion (from oxygen atom of water). This is called ligand to metal charge transfer. Any electron transfer is associated with absorption of light. This absorption process happens in the red end of the spectrum, leaving blue colour out of white light. That's why hydrated CuSO4 is blue. There is no such possibility in absence of water molecules, so anhydrous CuSO4 is white, meaning there is no charge transfer, no absorption and hence no colour.
The blue color of copper(II) sulfate is due to the d-d transition for the d9 configuration of Cu(II) and has nothing to do with LMCT.What happens to this transition after the sulfate is dried?
This question has been thoroughly answered in the following article:
Wildner, M., Giester, G., Kersten, M. et al. Polarized electronic absorption spectra of colourless chalcocyanite, CuSO4, with a survey on crystal fields in Cu2+ minerals. Phys Chem Minerals 41, 669–680 (2014). https://doi-org/10.1007/s00269-014-0681-1
In essence, the ligand field of the "4+2" coordination of the d(9) electron configuration by six sulfate ions in the anhydrate is sufficiently weaker than that of the four water + 2 sulfate coordination in CuSO4.5H2O, that the absorption band in the anhydride is shifted completely into the infra-red (lower energy) region. Even in the hydrate, the majority of the absorption is also in the infrared with just the band tail extending into the red region of the visible. For a spectral plot see the answer by Franz Wimmer on Quora: https://www.quora.com/What-makes-anhydrous-CuSO4-turn-blue-in-color