Find a polynomial P_{n} \in C[X] such as S_{n}=ker P_{n} (D) With S_{n}(n \geq 1)

Let me just try to improve the formulation of the question, which is quite confused. Before talking about endomorphisms, one should state which vector space one is considering; apparently this is the space ${\displaystyle E=C^{\mathrm{\infty }}\left(R\right)}$ of smooth functions on R (with values in the field we are working over, which I presume is C or maybe R).
Now the operation D of differentiation is an endomorphism of E (because it is a linear operation, and also because differentiating a ${\displaystyle C^{\mathrm{\infty }}}$ function results in another ${\displaystyle C^{\mathrm{\infty }}}$ function; note that the latter would not have worked for any finite differentiability class). Composition is a (bilinear) operation on endomorphisms, so ${\displaystyle D\circ D}$ (differentiation twice) is another endomorphism, and so is the k-fold composition ${\displaystyle D^k}$ for any ${\displaystyle k\in N}$; the latter is written ${\displaystyle D^k:f\to f^{\left(k\right)}}$ on the level of individual vectors ($C^{\mathrm{\infty }}$-functions) ${\displaystyle f\in E}$. (For ${\displaystyle k=0\text{one has}{f}^{\left(0\right)}=f}$.)
Since the set of endomorphism is also a vector space, we can form (finite!) linear combinations of these ${\displaystyle D^k}$ for different values of k with coefficients ${\displaystyle c_k}$ to get a quite general d-th order differential operator ${\displaystyle \sum _{k=0}^d{c}_k{D}^k}$ that maps ${\displaystyle f\to \sum _{k=0}^d{c}_k{f}^{\left(k\right)}}$. It is natural to consider this in relation to the polynomial ${\displaystyle P=\sum _{k=0}^d{c}_k{X}^k}$, namely as the result of substituting D for X into P. I will then write ${\displaystyle \sum _{k=0}^d{c}_k{D}^k=P[X=D]}$, which (since X is the only candidate to substitute for) can be shortened to P[D] (but I refuse to write this as P(D) as almost everybody does, since that would suggest that P is used as a function, which it is not). Finally, like for any endomorphism, one can form ${\displaystyle ker\left(P\left[D\right]\right)}$ to denote the subspace ${\displaystyle \{f\in E\mid P\left[D\right]\left(f\right)=0\}=\{f\in E\mid \sum _{k=0}^d{c}_k{f}^{\left(k\right)}=0\}}$ of vectors that are set to the zero vector by P[D].
Now the question becomes for which polynomial P this kernel gives the set of solutions to the differential equation ${\displaystyle f^{\left(n\right)}=f}$. It should be immediately obvious that ${\displaystyle P=X^n-1}$ is the unique polynomial that satisfies the requirement.