$V$ be a complex i.p.s. of dimension $n$ , $f(x) \in \mathcal P_{n-1}(\mathbb C)$ then $\exists $ linear operator $T$ on $V$ s.t. $T^*=f(T)$? Let $V$ be a complex inner product space of dimension $n$ , let $f(x)$ be a polynomial with complex-coefficients of degree less than or equal to $n-1$ , then is it true that there exist a linear operator $T$ on $V$ such that $T^*=f(T)$ ? ( note that such an operator $T$ must be normal ) 
 A: Since, as you observed, our search can be restricted to normal $T$'s, we can diagonalize $T$, and then all eigenvalues must satisfy
$$
f(z)=\overline{z} , \quad\quad\quad\quad\quad\quad (1)
$$
so existence of a $T$ is equivalent to the solvability (1).
This equation always has solutions (the "counterexample" I gave in an earlier version of this answer was based on a miscalculation). More generally, a polynomial of the form $z^n+p(x,y)$, $\deg p\le n-1$ (but not necessarily holomorphic), will have a zero. This proof of the fundamental theorem of algebra still works in this more general setting.
Update: In view of the obvious contradiction to the other answer, it's perhaps worth pointing out that (as I now realize, though it's actually obvious from the second paragraph) this is correct for $n\ge 2$.
A: The answer is no. Take for example the polynomial $f(x)=x+1$. Then there is no linear transformation $T$ on any $n$ dimensional complex inner product space $V$, (with $n\ge 2$), satisfying $T^*=T+I$. because if such a $T$ exists then taking the trace shows that $n=-2 i \Im({\rm tr} T) \in i\mathbb{R}$, a contradiction.
