In the normal case, we know that if an $n-$dimensional random variable $X$ has a multivariate normal distribution with mean vector $\mu\in \mathbb{R}^n$ and covariance matrix $C\in \mathbb{R}^n \times \mathbb{R}^n$, i.e., $X\sim \mathcal{N}(\mu,C)$, then $$ Z=C^{-1/2}(X-\mu)\sim\mathcal{N}(0,I_n), $$ that is, every component r.v. $Z_i$ of the vector $Z$ now has a standard normal distribution and each $Z_i$ is independent of $Z_j$, for $i\neq j$.
I just want to ask: is there already a result like this for other (multivariate) distributions? For example, given $X_1$ and $X_2$ form a bivariate exponential distribution with covariance $C$, can we form random variables $Y_1, Y_2$, still forming a bivariate exponential distribution but are now independent?
Probably not. Covariances don't do anything comprehensible under nonlinear transformations, and the exponential distributions aren't closed under linear ones that mix two or more of them. Further, all you could hypothetically do by considering covariances is prove you get uncorrelated variables, and Normal distributions are the only ones for which a multivariate distribution having uncorrelated components ensures their independence. In fact, I doubt the covariance matrix of a bivariate exponential distribution of known mean uniquely determines what the joint distribution is. Normal distributions are very special, because of the Pythagorean properties of the log-PDF.