We cannot work with multivariate normal distribution unless all variates follow normal distribution. Can we somehow transform all to normal?
I very much like the way you are thinking, but the answer is no. Given a sample space $E$. Any probability distribution $X$ over the sample space $E$ with probability measure $\mu_x$ can be transformed into any other probability distribution $Y$ over the sample space $E$ with probability measure $\mu_y$ as long as the probability measures are absolutely continuous.
Example 1: You cannot transform a discrete distribution (like the Poisson distribution) into a normal distribution because the Poisson distribution only has non-zero probability on the natural numbers whereas a 1 dimensional normal distribution has non-zero probability over the entire real line.
In addition, under certain circumstances, a distribution $X$ over sample space $E$ with probability measure $\mu_x$ can be transformed into a distribution $Y$ over sample space $F$ and probability measure $\mu_y$ so long as there is a topological equivalence between the two spaces $E$ and $F$ such that the probability measures $\mu_x$ and $\mu_y$ can be made absolutely continuous. For example, the positive real line $(0, \infty)$ (note I am excluding zero) can be transformed to the entire real line $(-\infty, \infty)$ using the log transform which is a continuous bijective mapping between the two spaces.
Example 2: Think about how the log-normal distribution can be transformed to the normal distribution by taking the log-transform of a log-normal random variable. Note however that the positive real line including zero $[0, \infty)$ is not topologically equivalent to the entire real line $(-\infty, \infty)$ due to the closed boundary at zero and as such something like a truncated normal defined over $[0, \infty)$ cannot be transformed to a normal distribution over $(-\infty, \infty)$.
To summarize this: The two distribution need to have zero and non-zero probability in the same places or there needs to be a continuous bijective mapping between the sample spaces such that the two distribution can be made to have zero and non-zero probability in the same places.
For an intuitive background on the measure theory needed to understand this, check out a recent blog post: Measure Theory Made Ridiculously Simple. In particular, at the end of the post I show how a transformation of variables is actually a change of measure; however, there to simplify my treatment I did not discuss the requirement of topological equivalency between sample spaces.
Update Based on Comments: It has been brought to my attention that the question may be referring to the existence of transformations that operate on each variate separately. In such cases in addition to the existence of a continuous bijective mapping between two sample spaces $E$ and $F$ (where $F = R^n$ as the poster is discussing multivariate normality) there is an added requirement on the mapping. For notational purposes lets assume the question refers to a starting multivariate distribution $X=(x_1, \dots, x_n)$ over a sample space $E^n = E_1\times\dots\times E_n$.
We now have the following restrictions on the transformation (mapping):
- As before we require that there exists a bijective continuous mapping between $E$ and $R^n$ (This can trivially be the identity map if $E = R^n$).
- As before we require that this mapping is such that $\mu_x$ (the probability measure of the starting distribution) is absolutely continuous with a transformed measure $\mu_y$ (the gaussian measure over $R^n$.
- We now also require that the mapping has the form $f(x_1, \dots, x_n)=f_1(x_1)\circ\dots\circ f_n(x_n)$ such that $f:E^n \rightarrow F^n$ and where $f_i : E^i \rightarrow F^i$ such that the transform is essentially operating on each variate separately.
Adding this further restriction simply solidifies the conclusion that such a transformation may not always be possible.