For an odd integer $k \geq 1$, let $F$ be the set of all entire functions $f$ such that $$f(x)= |x^k|$$ for all $x \in (-1,1)$. Then the cardinality of $F$ is
$0$
$1$
$> 1$ but finite
Infinite.
I think for all integers $k \geq 3$, $f$ is entire. But the answer given is $0$.
If the function $f$ is an entire function which respects your hypothesis, it coincides with the function $g(z)=z^k$ on the set $(0,1)$ which is itself an entire function.
The set $(0,1)$ has at least one limit point, so by the analytic continuation principle, $f(z)=g(z)$ for all $z\in \mathbb{C}$.
However, the function $f$ is even when restricted to $(-1,1)$, and $g$ is an odd function if $k$ is odd.
The only function which is even and odd is the constant function which returns zero. This means that the function $f$ is not analytic.