In acidic solutions ($\mathrm{pH} = 0$), the values of the relevant reduction potentials of $\ce{Pu^3+/Pu^4+}$ ($\pu{+1.01 V}$), $\ce{Pu^3+/PuO2+}$ ($\pu{+1.03 V}$), $\ce{Pu^3+/PuO2^2+}$ ($\pu{+1.02 V}$), $\ce{Pu^4+/PuO2+}$ ($\pu{+1.04 V}$), $\ce{Pu^4+/PuO2^2+}$ ($\pu{+1.03 V}$), and $\ce{PuO2+/PuO2^2+}$ ($\pu{+1.02 V}$) are all very similar.
Therefore, plutonium has the remarkable feature that all four important oxidation states can be present simultaneously in aqueous solutions.
In aqueous solution, $\ce{Pu^4+}$ disproportionates into a mixture of oxidation states:
$$\ce{2Pu^4+ + 2H2O <=> Pu^3+ + PuO2+ + 4H+}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$
Thus, the net reaction of the disproportionation of $\ce{Pu^4+}$ is approximately:
$$\ce{3Pu^4+ + 2H2O <=> 2Pu^3+ + PuO2^2+ + 4H+}$$
Nevertheless, $\ce{Pu^4+}$ may still be the predominant oxidation state in the mixture.
Besides, the equations imply that the disproportionation of $\ce{Pu^4+}$ dcreseases as the acidity of the solution further increases.
Concerning pentavalent plutonium ($\ce{PuO2+}$), your expectations are correct. In typical acidic solutions, $\ce{PuO2+}$ indeed quickly disproportionates into a mixture of oxidation states:
$$\ce{2PuO2+ + 4H+ <=> Pu^4+ + PuO2^2+ + 2H2O}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$
However, the given values for the reduction potentials apply to pH = 0, and the various reduction potentials of plutonium depend on pH. Furthermore, the stability of individual oxidation states can be strongly influenced by hydrolysis, or the formation of complexes or compounds with low solubility products. Therefore, the predominant oxidation state can be changed by means of multiple factors. Furthermore, the valence of plutonium solutions may change as a consequence of radiolysis.
