Given a univariate sample $\vec X = X_1, ..., X_n$ with standard deviation 1 and a strictly monotone transformation $t: R \to R$ with the property that the standard deviation of $t(\vec X)$ is also 1 (where $t(\vec X)$ is $t$ applied to each $X_i$). If one know fits a normal distribution to $\vec X$ and $t(\vec X)$, one observes the following fact: the likelihood at the respective MLEs of $\vec X$ and $t(\vec X)$ is the same. The reason for that is, that the standard deviation is the MLE for the $\sigma$ parameter of the normal distribution, so just the mean changes, which does not change the likelihood.
In some computations with multivariate data I observed the same fact, using transformations such that $\det(cov(t(\vec X)))=1)$. But I did not see a straightforward proof for that! Note that in the multivariate case, the MLEs for the variance parameters do change, even if the above determinant is always 1.
I am sure I am not the first to notice, but my (somewhat limited) literature on the multivariate normal distribution does not give me a clue how to prove the above statement, i.e. that the multivariate normal likelihood does only depend on the determinant of the $cov(t(\vec X))$, regardless of choice of $t$.
A proof boils down to looking at the terms of the form $x'\Sigma x$ for $\det \Sigma = 1$.
Do you have more clue?