Let's study the possible roots of $f(x)$. If $c$ is a zero of $f(x)$, i.e. $f(c)=0$, then $f(c^2)=f(c)f(c+1)=0$ which means $c^2$ will be a root too. However, the number of zeros of a polynomial is limited. Therefore, $c=0$ or $\pm 1$. Meanwhile, if $c$ is a root, then $f((c-1)^2)=f(c-1)f(c)=0$ which means that $(c-1)^2$ will be a root too. Therefore, $c$ cannot be $-1$. Hence, $f(x)$'s root can only be $0$ or $1$. It follows that $$f(x)=kx^m(x-1)^n$$ where $k$, $m$ and $n$ are non-negative integers. Setting this to the give relation yields \begin{align} k(x^2)^{m}(x^2-1)^n &= \Big(kx^m(x-1)^n\Big)\Big((x+1)^mx^n\Big)\\ kx^{2m}(x^2-1)^n &=k^2x^{m+n}(x-1)^n(x+1)^m\\ kx^m(x+1)^n&=k^2x^n(x+1)^m \end{align} In order for this relation holds for any $x$, constant $k$ on the left must equal $k^2$ on the right. Hence, $k=0, 1$. If $k=0$, we find $f(x)=0$ is one solution. If $k=1$. $f(x)=x^m(x-1)^n$ and $x^m(x+1)^n=x^n(x+1)^m$ hold for any $x$. If $m\ne n$, without lose of generality, let's assume $m > n$. Then $x^{m-n}=(x+1)^{m-n}$. This cannot hold. Therefore $m$ and $n$ must equal. Consequently, $f(x)=x^n(1-x)^n$. We then conclusion there are totally two solutions: $\boxed{f(x)=0}$ and $\boxed{f(x)=x^n(x-1)^n}$ where $n$ is a non-negative integer.