P(x²) = P(x)²

Answer. $P=0$ and $P(x)=x^n$ for $n\ge0$ (including $P=1$).

These work. $P(x^2)=x^{2n}=(x^n)^2$.

Converse. Let $P\ne0$ have degree $n$ and leading coefficient $a$. Comparing the coefficients of $x^{2n}$: $a=a^2$, so $a=1$.

Suppose $P$ has another nonzero term, and let $bx^j$ ($b\ne0$, $j<n$) be the highest of the remaining terms. Look at the coefficient of $x^{n+j}$.

*In $P(x)^2$:* it is a sum of $c_ic_{i'}$ over pairs of exponents $(i,i')$ with $i+i'=n+j$ and both coefficients nonzero. If neither exponent is $n$, both are $\le j$, and $i+i'\le2j<n+j$: impossible. So one exponent is $n$ and the other is $j$, and the coefficient is $2\cdot1\cdot b=2b\ne0$.

*In $P(x^2)$:* only even exponents $2e$ appear, where $e$ is an exponent of $P$. For $x^{n+j}$ we would need $e=\frac{n+j}2$, which lies strictly between $j$ and $n$, and $P$ has no nonzero term there. So the coefficient is $0$.

The two sides differ: contradiction. Hence $P$ has a single nonzero term, which is $x^n$. $\blacksquare$

Lesson. When comparing coefficients in a polynomial identity, look at the leading term and the *next* term: one of the two sides often cannot produce the exponent the other one does.