Existence of a map $\phi:\mathbb{Z}_{N^2}^* \mapsto \mathbb{F} $ Is there a map between the group of $\mathbb{Z}_{N^2}^*$ where $N$ is a composite number , a product of two equal size secure prime numbers $p$ and $q$ and a finite field $\mathbb{F}$, such that for all $x \in \mathbb{Z}_{N^2}^*$ there is a polynomial representation of $x \in \mathbb{F}$ ?
So if you want a homomorphism from $\mathbb{Z}_{N^2}^*$ into the multiplicative group of a field, you have to adjust the number of elements in the field. With $p=2a+1$ and $q=2b+1$ (with $a,b$ prime), you have
$$|\mathbb{Z}_{N^2}^*| = 4pqab$$
The multiplicative subgroup of a field $F_{x^y}$ with prime $x$ and exponent $y$ has $x^y-1$ elements. At this point you could choose a specific $N$, that the according cardinalities match ($4abpq=x^y-1$), but if you have to apply this to arbitrary $N$, it won't work this way. Then all you can do is try to find $x^y$, s.t. $(x^y-1)$ is a factor of $4abpq$. If you use e.g. $\mathbb{F}_p^*$ or $\mathbb{F}_q^*$, this will work. But I can't think of any other fields which have this property from these given values.

As an example for an homomorphism you can use $f(x) = x^{pq(q-1)}$ to get a subgroup of order $p-1$, which you might be able to find an isomorphism into the multiplicative group of $\mathbb{F}_p$.

However, a bijective function (isomorphism) is quite unlikely to exist.