Any typo or I have not understood anything: a) The operation + is not associative: $(a+b)+c=b+c=a\ne 0 =a+a=a+(b+c)$. So what you mean by $x+y+z$? b) $f(x+x)=f(0), \ \ f(x)+f(x)=0$ So $I$ consists exactly of functions for which $f(0)=0$ Hence trivially we have $|I|=4^{3}$

digger wrote: a) The operation + is not associative: $(a+b)+c=b+c=a\ne 0 =a+a=a+(b+c)$. So what you mean by $x+y+z$? I edited

$X$ is just the $\mathbb{F}_{2}$-vector space $\mathbb{F}_{2}^{\ 2}=\mathbb{F}_{2}\times \mathbb{F}_{2}$ with its addition. So $f(x+y+x)=f(y)=f(x)+f(y)+f(x)$ for all $f: X\rightarrow X$ and $x,y\in X$. So $S=M(X)=X^{X}$ and $|S|=|X|^{|X|}=4^{4}$.