Polynomial equation

Let $P \in \mathbb{C}[X]$ such that $P(X^2) = P(X)^2$

1. If $P=c \in \mathbb{R}$
   $$c = c^2 \implies c(c - 1) = 0 \implies c \in \{0, 1\}$$
2. If $P$ is not constant, let $n=\deg(P)$
   We find the leading coefficient of $P$:
   Let $a_n$ be the leading coefficient of $P(X)$. By comparing the leading coefficients on both sides of $P(X^2) = P(X)^2$:
   $$a_n = a_n^{2} \implies a_n \in \{0, 1\}$$Since $\deg(P)=n$, $a_{n} \neq 0$.
   Therefore, $a_{n}=1$

With the leading coefficient known, we can write, using the Euclidean division of $P(X)$ by $X^n$:
$$P(X) = X^n + R(X) \quad \text{where } \deg(R) = q < n$$Substituting this into $P(X^2) = P(X)^2$:
$$X^{2n} + R(X^2) = \left(X^n + R(X)\right)^2 = X^{2n} + 2X^n R(X) + R(X)^2$$Subtracting $X^{2n}$ from both sides gives:
$$R(X^2) = 2X^n R(X) + R(X)^2$$Now, assume $R \neq 0$: The left-hand side has degree $\deg\left(R(X^2)\right) = 2q$. On the right-hand side, $\deg\left(2X^n R(X)\right) = n + q$ and $\deg\left(R(X)^2\right) = 2q$. Since $q < n$, we have $2q < n + q$, so the degree of the right-hand side is strictly $n + q$. Equating the degrees:
$$2q = n + q \implies q = n$$This contradicts the condition $q < n$. Thus, we must have $R = 0$, which implies $P(X) = X^n$.

Conversely, if $P=X^n$, $P(X^2)=X^{2n}=(X^{n})^{2}=(P(X))^{2}$

Conclusion: The set of all polynomial solutions in $\mathbb{C}[X]$ is:
$$\mathcal{S} = \{0\} \cup \{X^n \mid n \in \mathbb{N}\}$$