I was wondering if there is a continuous function $f:[0,1]\to\mathbb R$ satisfying $$f(x)^2=f\left(x^2\right)\text,$$ for all $x\in[0,1]$, $f(0)=1$ and $ f(1)=0$.
Clearly, some easy functions like polynomials are not satisfied. I guess there is a way to construct an example since it only needs a continuous function.
There is no such function. Let $g(x)=|f(x)|$. Note that $g(x^{2})=(g(x))^{2}$. By iteration we get $g(x^{2^{n}})=(g(x))^{2^{n}}$ for every positive integer $n$. By continuity $g(x^{2^{n}}) \to g(0)=1$ for $|x| <1$. So $(g(x))^{2^{n}} \to 1$ for such $x$. But this implies $g(x)=1$. [Indeed $g(x) >1$ implies $(g(x))^{2^{n}} \to \infty$ and $g(x)<1$ implies $(g(x))^{2^{n}} \to 0$]. We have proved that $|f(x)|=1$ for $|x|<1$. By continuity $|f(1)|=1$ contradicting the hypothesis.