Oxidation of water by permanganate: Does this reaction happen?
You have indeed gotten it terribly wrong. The key hint is that you have more than one element that would be reduced: hydrogen also goes from $\mathrm{+I}$ to $\pm0$ in your sample equation. Furthermore, in neutral or weakly basic solutions permanganate gives $\ce{MnO2}$ after oxidation, not manganese(III).

What you do want to do is to reduce permanganate to something (which I have already established would be manganese dioxide) while oxidising something in water to something. Looking at the oxidation states of hydrogen and oxygen in water, the only element that can be oxidised is oxygen, since hydrogen is already at its highest possible oxidation state $\mathrm{+I}$. Oxidising water’s oxygen leads to a few possibilities of which oxygen gas does indeed sound most likely. Therefore, we have the redox pairs:

$$\begin{align}\text{Red: }\ce{\overset{+VII}{Mn}O4-}/\ce{\overset{+IV}{Mn}O2}&amp;&amp;\text{Ox: }\ce{H2\overset{-II}{O}}/\ce{\overset{\pm0}{O2}}\end{align}$$

We can then determine the appropriate half-reactions by the standard procedure. Instead of walking you through it, I will present you with the final result for comparison.

$$\begin{align}\ce{MnO4- + 3e- + 4 H+ &amp;-&gt; MnO2 + 2 H2O}\tag{Red}\\
\ce{2H2O &amp;-&gt; O2 + 4e- + 4 H+}\tag{Ox}\\[1em]\hline
\ce{4 MnO4- + 4 H+ &amp;-&gt; 4 MnO2 + 3O2 + 2 H2O}\tag{Redox}\end{align}$$

It turns out that the final reaction equation formally doesn’t even need water. However, potassium permanganate is a shelf-stable solid under normal laboratory conditions (if light is excluded), so it is reasonable to assume that the reaction will only proceed in aqueous solution.

To determine whether the redox reaction is feasible in aqueous solution, it is easier to go with the two half-reactions we already have; thermodynamically, they must correspond to the same process with the same reaction enthalpy etc. The Nernst equation together with the inspection of the cell potential allows us to estimate whether a process will be spontaneous or not by mere inspection of the standard potentials.

$$\begin{align}E^0_\text{cell} &amp;= E^0_\text{Red} - E^0_\text{Ox}\\
&amp;= \pu{+1.70V} - (\pu{+1.23V})\\
&amp;= \pu{+0.47V}\\
E^0_\text{cell} &amp;&gt; 0 \\&amp;\Longrightarrow \text{spontaneous process}\end{align}$$

Thus, thermodynamics already predict the decomposition of permanganate in aqueous solution to be spontaneous under standard conditions. As mentioned, the Nernst equation will allow us to estimate whether it also is spontaneous given general laboratory conditions; I will assume $\mathrm{pH} = 7$ and $p(\ce{O2}) = \pu{0.2bar}$

$$\begin{align}E &amp;= E^0 - \frac{0.059}{z} \lg \frac{Red}{Ox}\\[0.5em]
E &amp;= \pu{+0.47V} - \frac{\pu{0.059V}}{12}\lg\frac{p(\ce{O2})^3}{[\ce{MnO4-}]^4[\ce{H+}]^4}\\[0.5em]
E &amp;= \pu{+0.47V} - \frac{\pu{0.059V}\times28\times2.1\times(-4)}{12}\lg {[\ce{MnO4-}]}\\[0.3em]
E &amp;= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\end{align}$$

We can calculate the minimum concentration from which the process will be spontaneous:

$$\begin{align}\pu{0V} &amp;= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\\
\pu{-0.47V} &amp;= \pu{1.16V}\times\lg [\ce{MnO4-}]\\
-0.405 &amp;= \lg [\ce{MnO4-}]\\
10^{-0.405} &amp;= [\ce{MnO4-}]\\
0.393 &amp;= [\ce{MnO4-}]\end{align}$$

Thus, at any concentration higher than $\pu{0.393M}$ permanganate should start formally oxidising water or decomposing.

Whether or not this happens at exactly this concentration depends on kinetic factors, i.e. can oxygen actually be produced under these conditions, but thermodynamically it would be favourable.