"Is it possible to have a homomorphic mapping from $\mathbb F_{p^n}$
to $\mathbb Z_{p^n}$ that preserves both the add and multiplication operators?"
Other than the isomorphism when $n=1$, only the very boring mapping that send everything to 0. Consider the multiplicative identity of $\mathbb F_{p^n}$. We write this as 1 and consider our putative homomorphism $\phi$. We see that by additivity adding $k$ copies of 1, for any integer $k$ we have $\phi(1+\cdots+1)=k\phi(1)\mod {p^n}$ and in particular with $k=p$ we see that $p\phi(1)=\phi(0)=0$ so that $\phi(1)=cp^{n-1}$ for some $1\le c\le p$. Moreover, by multiplicativity we have $\phi(1)=\phi(1\cdot 1)=\phi(1)\phi(1)$ so that $\phi(1)=1$ or $0$. We conclude that $c=p$ and $\phi(1)=0$ (except in the case $n=1$). Further, for any $\alpha\in\mathbb F_{p^n}$ $\phi(\alpha)=\phi(1\cdot\alpha)=\phi(1)\phi(\alpha)=0$.
"Or if we relax requirement, can we have a homomorphic mapping from the multiplicative group $\mathbb F_{p^n}^\times$ to $\mathbb Z_{p^n}^\times$ which preserves multiplication?"
Only marginally less boring. Note that $|\mathbb F_{p^n}^\times|=p^n-1$ and $|\mathbb Z_{p^n}^\times|=p^n-p^{n-1}$. The size of the image of any homomorphism has to divide the GCD of these two which is $p-1$. We see that the image has to be a subgroup of the $(p-1)$th roots of 1 in $\mathbb Z_{p^n}$. Now pick any multiplicative generator of $\mathbb F_{p^n}^\times$ call this $\alpha$. There are precisely $p-1$ group homomorphisms depending on which of the $(p-1)$th roots 1 is equal to $\phi(\alpha)$. The kernel will be the $\ell$th powers in $\mathbb F_{p^n}^\times$ where $\ell|(p-1)$ is the multiplicative order of $\phi(\alpha)$ in $\mathbb Z_{p^n}^\times$.
