When doing intertheoretic utility comparisons, there is one clear and easy case: when everything is exactly symmetric.

This happens when, for instance, there exists $u$ and $v$ such that $p\left(\right[u\left]\right)=p\left(\right[v\left]\right)=0.5$ and there exists a map $\sigma :S\to S$, such that ${\sigma }^2$ is the identity (hence $\sigma$ is an involution) and for all $s\in S$), $u\left(s\right)=v\left(\sigma \right(s\left)\right)$.

Note that this implies that $\left(u\right(s),v(s\left)\right)=\left(v\right(\sigma \left(s\right)),u(\sigma \left(s\right))$, so $\sigma$ is essentially a 'reflection'.

Then, since everything is so symmetric, we can say there is no way of distinguishing $u$ from $v$, so the correct approach is to maximise $[u+v]$.

See the following graph, with the strategy to be followed marked in red:

Permutation

Symmetry is good as far as it goes, but is very fragile. It doesn't say anything about what happens when $p\left(\right[u\left]\right)=49/100$ and $p\left(\right[v\left]\right)=51/100$, for instance.

There is an argument, however, that resolves the unequal probability cases and extends the results to non-symmetric cases.

Consider for instance the case where $u$ and $v$ are as follows, for $5$ strategies in $S$:

Nothing obvious springs to mind as to what the best normalisation process is -- the setup is clearly unsymmetrical, and four of the five options are on the Pareto boundary. But let's rescale and translate the utilities:

This is still unsymmetrical, but note that $u$ and $v$ have the same values: $-1$, $0.5$, $2$, $3$, and $3.5$.

Thus there is a permutation $\rho :S\to S$ such that for all $s\in S$, $u\left(s\right)=v\left(\rho \right(s\left)\right)$.

Another type of uncertainty

This permutation $\rho$ allows us to transform the uncertainty about one's own values (which we don't know how to handle) to other types of uncertainty (which we can).

How so? Let $S^{\prime }$ be another copy of $S$, but with different labels, and let $i:S\to S^{\prime }$ be the identity map that re-assigns each element to its original label.

Then instead of seeing $u$ and $v$ as different utility functions on $S$, we can see them as both being the same utility function $w$ on $S^{\prime }$, with uncertainty over a map $m$ from $S$ to $S^{\prime }$. This map $m$ is $i$ with probability $p\left(\right[u\left]\right)$ and $i\circ \rho$ with probability $p\left(\right[v\left]\right)$.

Thus the agent should see itself as a $w$-maximiser, with standard uncertainty over the map $m$ (this could be seen as uncertainty over the consequences of choosing a strategy). This means it will maximise $\left[p\right(\left[u\right])u+p(\left[v\right]\left)v\right]$, as long as $u$ and $v$ are related by a permutation on $S$.

In this particular case, if $p\left(\right[u\left]\right)$ and $p\left(\right[v\left]\right)$ are close to $0.5$, this means that the red strategy will be selected:

Note that permutations includes the symmetric case where $\rho$ is an involution, so the symmetric argument now extends to cases where $p\left(\right[u\left]\right)\ne p\left(\right[v\left]\right)$.

Consequences

For permutations, this argument claims that the correct approach is to use Individual normalisations. Indeed, every normalisation presented here would reach the same result.

This doesn't prove that individual normalisation is necessarily correct -- you can imagine a general system that only give individual normalisations on permutation problems -- but is suggestive none the less.