Alias Structure of one-fourth replicate of a $4^2$-Factorial Design with interaction $\text{A}{B}^3$ confounded For finding the Alias of main effect A , i started as the following :
$\text{A}$$\times$ $\text{A}{B}^3$$=\text{A}^2\text{B}^3={(\text{A}^2\text{B}^3})^2=\text{A}^4\text{B}^6=\text{B}^2$(mod $4$)$={(\text{B}^2)}^2=\text{B}^4=0$(mod $4$)
Where am i doing mistake?
 A: I strongly recommend reviewing Wu and Hamada (2011) for a full treatment of $q^k$ factorials and their fractions when $q > 2$. 
First, note that $A^2B^3 = (A^2B^3)^2$ only when $A^2B^3 = I$. Do you know this to be true?  Does $B^2=I$ as well?  
The second issue is a bit more involved.  You're using codeword notation, but we can write these words in terms of the finite field: $A^2B^3$ would be $2A+3B$.  Restating part of your expression above, we may write $$(2A+3B)+(2A+3B) \equiv 2B \mod 4.$$ It's more convenient to use this approach when $q > 2$. You then have a concern for how to convert the levels of $\mathbb{F}_q$ to levels in the experiment.  Again, look to Wu and Hamada (2011).
It is a wonderful coincidence that, for $2^k$ designs and their fractions with levels $\{-1,1\}$, codewords have the same values as the contrasts of the same name in $\mathbb{R}$.    For example, the word $AB$, or $A+B$ in $\mathbb{F}_2$, when converted $\{-1,1\}$ is identical to the contrast for $AB$ in the linear model.  
This is not generally true for $q^k$ designs.  Even if the words $A=AB^3$, or $A = A + 3B$ in $\mathbb{F}_4$, that doesn't tell you about the relationship between the contrasts for $A$ and $AB^3$.  You need to look at the alias matrix which is also described in Wu and Hamada (2011).
References
Wu, J. and Hamada, M. (2011). Experiments: Planning, Analysis, and Optimization. 2ed. John Wiley & Sons.
