Let $f \colon [0,1] \to \mathbb R$ be continuous and let $f_n$ be defined by $$ f_n(x):=f(x^n), \qquad x \in [0,1], \,\, n \in \mathbb N. $$ Suppose there exists a continuous function $g$ on $[0,1]$ such that $f_n \to g$ uniformly. Then $f$ is constant.
Let $\overline{x} \in [0,1)$ be fixed. Then we have that the sequence of real numbers $x_n:=\overline{x}^{n} \to 0$ as $n \to +\infty$.
On the one hand, by continuity, we have
$$
f(x_n) \to f(0);
$$
on the other hand, being $f(x_n) = f(\overline{x}^{n}) = f_n(\overline{x})$, by pointwise convergence, we deduce
$$
f(x_n) = f_n(\overline{x}) \to g(\overline{x}).
$$
Hence for every $x \in [0,1)$ we have $g(x)= f(0)$.
If $x=1$ then we have $f(1)=f_n(1) \to g(1)$. By continuity, $g$ must be constant and in particular we have $f(0)=f(1)$.
Now for any $\overline{x} \in (0,1)$ we have $$ f(\overline{x}) = f_n(\overline{x}^{1/n}) \to g(1) $$ because $f_n \to g$ uniformly and $\overline{x}^{1/n} \to 1$. Hence we must have $f(x)=g(1)$ for any $x \in (0,1)$.
Thus we deduce that $f$ is constant on $[0,1]$.
Is this correct? Thanks.