Find a linear functional on $M_n(\mathbb{C})$ which preserves matrix multiplication

There is only one multiplicative functional $f: M_n(\mathbb{C}) \to \mathbb{C}$ when $n > 1$, which is $f=0$.

Indeed, if $f: M_n(\mathbb{C})\to \mathbb{C}$ is a multiplicative functional, then $\ker(f)$ is a two-sided ideal in $M_n(\mathbb{C})$. But $M_n(\mathbb{C})$ is a simple algebra, so either $\ker(f) = M_n(\mathbb{C})$ which implies $f=0$ or $\ker(f)=0$ which implies $f$ is injective, which is impossible by dimension reasons.

Another proof which doesn't need simpleness of $M_n(\mathbb{C})$:

Let $E_i$ be the identity matrix with all columns set to zero except for the $i$th. Then $f(E_i) f(E_j) = 0$ whenever $i \not = j$, which means there is at most one $k$ with $f(E_k)$ nonzero. But also $f(Q X Q^{-1}) = f(X)$ for all $X$ and invertible $Q$, and all of the $E_i$s are similar, so $f(E_k) = 0$ too (if $n > 1$). Finally then for any $X$ we have $f(X) = f(X I) = f(X) \sum_i f(E_i) = 0$, as desired.