[Dolphinsiu]

[V(O)(H2O)5]SO4 is d1 complex as V(0) is d5 complex. Six coordination of this complex, giving rise to 12 electrons, is hence coordinatively unsaturated. V(IV) is a very high oxidation state. Hence it is electron-deficient metal center so that H2O in [V(O)(H2O)5]2+ is highly acidic. Upon addition of water, deprotonation of metal complex exists with release of proton.

[V(O)(H2O)5]2+ + H2O --> [V(O)(H2O)4(OH)] + + H3O+

In above reaction, HSO4- is the counter ion for the product.

[V(O)(H2O)4(OH)] + + H2O --> [V(O)(H2O)3(OH)2] + H3O+

[V(O)(H2O)3(OH)2] + + H2O --> H[V(O)(H2O)2(OH)3]
--> H2[V(O)(H2O)1(OH)4] --> H3[V(O)(OH)5]

Am I correct?

[KhemistKen]

I would expect that a small portion of the vanadium would react as in your first equation.  Wouldn't expect any significant contribution from the other reactions however.