Likelihood at MLE and transformations, the multivariate normal case 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?
Thanks, Philipp
 A: This is due to the invariance property of MLEs.  It is an elementary exercise in calculus to show that if $E_{MLE}$ maximises the function $F(E)$ then any monotonic function of $E$, say $g=g(E)$ will also be maximised at $g_{MLE}=g(E_{MLE})$.
$F(E)=F[g^{-1}(E)]$ and therefore the maximum occurs at $F_{MAX}=F(E_{MLE})=F[g^{-1}(E_{MLE})]$, which means that $g_{MLE}=g(E_{MLE})$.
If $g(.)$ were not monotone (1-to-1 so that $g^{-1}(.)$ is well defined), then the above argument would not hold.
EDIT: I think this is why the MLE's change for the determinant transformation, perhaps the $t(.)$ function is not a  1-to-1 function.  It is all about the functional properties of your $t(.)$ function (whether it is 1-to-1 or not).
EDIT#2: After some discussion, the transformation $t(.)$ may be 1-to-1, but if the inverse transformation $t^{-1}(.)$ doesn't have a Jacobian of 1, then the value of the likelihood at its maximum will be different.  Fixing the variance to be 1 may one way of achieving this property.
The class of transformations $t(X)$ has a approximate (exact for normals?) covariance of $$cov[t(X)]\approx (\frac{\partial t(x)}{\partial t})_{x=X_{MLE}}^T cov[X](\frac{\partial t(x)}{\partial t})_{x=X_{MLE}}$$.
Fixing the determinant to 1 fixes (approximately) 
$$|\frac{\partial t(x)}{\partial t}|^{-2}_{x=X_{MLE}}\approx |cov[X]|$$
Now if $t(.)$ is invertible, the jacobian of the inverse transformation is just the inverse of the jacobian.  SO we have:
$$|\frac{\partial t(x)}{\partial t}|^{-1}_{x=X_{MLE}}=|\frac{\partial t^{-1}(x)}{\partial t}|_{x=X_{MLE}}\approx |cov[X]|^{\frac{1}{2}}$$
Now in the normal distribution, the likelihood evaluated at the MLE, the exponential part just becomes 1.  So the maximised likelihood  for X is given by
$$f_X(x_{MLE}) = \frac{1}{(2\pi)^{\frac{N}{2}} |cov[X]|^{\frac{1}{2}}}$$
using the Jacobian rule for transforming to $T=t(X)$ gives
$$f_T(t_{MLE}) = \frac{|\frac{\partial t^{-1}(x)}{\partial t}|_{x=X_{MLE}}}{(2\pi)^{\frac{N}{2}} |cov(X)|} \approx\frac{|cov[X]|^{\frac{1}{2}}}{(2\pi)^{\frac{N}{2}} |cov[X]|^{\frac{1}{2}}} = (2\pi)^{-\frac{N}{2}}$$
Which is a constant, as your analysis shows (you could even check my calculations, by comparing the likelihood you get with $(2\pi)^{-\frac{N}{2}}$, and see if it is right).  I would imagine that either the approximation is exact for normals, or exact for certain transformations $t(.)$, or becomes exact as $N\rightarrow\infty$, or the approximation error is in decimal places which you checked (or can't calculate), OR the computer you use to get the MLEs of $T$ uses the same approximation technique as I have (called the delta method).
