Let's determine those polynomials $f(x)$ for which there exists a polynomial $p(t)$ that satisfies the identity

$$
f\left(x^{2}\right)=p(f(x))
$$
Let $f(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}$ be an $n$-th degree polynomial. Then $p(f(x))=f\left(x^{2}\right)=a_{n} x^{2 n}+a_{n-1} x^{2 n-2}+\cdots+a_{0}$ has degree $2n$. If the degree of the polynomial $p(t)$ is $z$, then the degree of $p(f(x))$ is $z \cdot n$, from which $z \cdot n=2n$. For $n \geq 1$, dividing by $n$ gives $z=2$. Thus, the polynomial $p(t)$ must be of the form $a t^{2}+b t+c$, where $a \neq 0$.

Suppose in the polynomial $f(x)$, $a_{n-1}=a_{n-2}=\cdots=a_{k+1}=0$, but $a_{k} \neq 0$ and $k \geq 1$:

$$
f(x)=a_{n} x^{n}+a_{k} x^{k}+\cdots+a_{0}
$$

The coefficient of $x^{n+k}$ in the polynomial $f^{2}(x)$ is $2 a_{n} a_{k}$, and in the polynomial $f(x)$ it is 0, because $f(x)$ is of degree $n$. Thus, in the polynomial $p(f(x))=a f^{2}(x)+b f(x)+c$, this coefficient is $2 a a_{n} a_{k}$. But $f^{2}(x)=a_{n} x^{2 n}+a_{k} x^{2 k}+\ldots$, and here the coefficient of $x^{n+k}$ is 0, because $2 n>n+k>2 k$. Then $0=2 a a_{n} a_{k}$, which contradicts the fact that the numbers $a, a_{n}, a_{k}$ are not zero.

Thus, for $n \geq 0$, the polynomial $f(x)$ can only be of the form $a_{n} x^{n}+a_{0}$. If $n=0$, then $f(x)=a_{0}$, and this polynomial always satisfies the problem if the polynomial $p(t)$ is chosen to be $a_{0}$.

We still need to examine the case when $f(x)=a_{n} x^{n}+a_{0}$, where $a_{n} \neq 0$. Then

$$
p(f(x))=a\left(a_{n} x^{n}+a_{0}\right)^{2}+b\left(a_{n} x^{n}+a_{0}\right)+c=a a_{n}^{2} x^{2 n}+\left(2 a a_{0} a_{n}+b a_{n}\right) x^{n}+a a_{0}^{2}+b a_{0}+c, f\left(x^{2}\right)=a_{n} x^{2 n}+a_{0}
$$

Thus, $f\left(x^{2}\right)=p(f(x))$ if and only if $a a_{n}^{2}=a_{n}, 2 a a_{0} a_{n}+b a_{n}=0, a a_{0}^{2}+b a_{0}+c=a_{0}$. Dividing the first equation by $a_{n}^{2}$ gives $a=\frac{1}{a_{n}}$. Substituting this into the second equation: $2 a_{0}+b a_{n}=0$, from which $b=-\frac{2 a_{0}}{a_{n}}$. Finally, let's look at the third equation:

$$
\frac{1}{a_{n}} a_{0}^{2}-\frac{2 a_{0}^{2}}{a_{n}}+c=a_{0}, \quad \text { from which } \quad c=\frac{a_{0}^{2}}{a_{n}}+a_{0} .
$$

Thus, for the polynomial

$$
p(t)=\frac{1}{a_{n}} t^{2}-\frac{2 a_{0}}{a_{n}} t+\frac{a_{0}^{2}}{a_{n}}+a_{0}
$$

we have $p(f(x))=f\left(x^{2}\right)$.

Therefore, the solution to the problem is:

$$
f(x)=a_{n} x^{n}+a_{0}, \quad \text { where } \quad n \geq 0 .
$$

Based on the work of Bence Fejérvári (Győr, Czuczor G. Bencés Gimn., IV. o.t.)