Do Legendre Transforms preserve the number of independent variables? Mathematical question about thermodynamics and natural variables/functions

The questions is to find a natural function of ${E, V,mu}$ in thermodynamics which is causing certain conceptual issues for me with the Legendre transform.

The function that has ${E,V, N}$ as natural variables is the entropy which is clear since the differential form of energy can be inverted for the differential form of entropy as the following:

$dE = TdS – PdV + mu dN $

$rightarrow dS = frac{1}{T} dE + frac{P}{T}dV -frac{mu}{T}dN$

Now, if I want to find the function that has natural variables of ${E, V, mu}$ which presumably seems easy as a Legendre transform from $Nrightarrow mu$.

Doing this in normal Legendre transform procedure, I SHOULD get the following:

$f(E,V,mu) = S(E,V,N) – frac{partial S}{partial N} N = S + frac{mu}{N}{T}$

The issue is that if I take the differential of $f$, I find that:

$df = frac{1}{T}dE + frac{P}{T}dV + frac{N}{T} dmu -frac{mu N}{T^{2}}dT$

In short, the pesky $dT$ term actually generates a new independent variable so that I didn’t get what was intended and instead got $f = f(E, V, T, mu)$. What is going on and why are more independent variables emerging than intended?

If I want to preserve 3 variables the best that can be done is $f = f(E, V, frac{mu}{T})$. Does this mean there isn’t actually a function with natural variables ${E, V,mu}?

I found these resources but they weren’t of much help.