Existence of an analytic function on disk

Clearly, constant $g\equiv 0$ or $g\equiv 1$ both satisfy this relation.

Assume $g$ is not constant.

If $g(z)=g^2(-z)$, then $g(0)=g^2(0)$, and hence $g(0)=0$ or $1$.

A. If $g(0)=0$, then $g(z)=z^kf(z)$, for some positive integer $k$ and $f$ analytic in the unit disk, with $f(0)=a\ne 0$. In such case we have $$ n^{-k}f(1/n)=g(1/n)=g^2(-1/n)=n^{-2k}f^2(-1/n) $$ and hence $$ f(1/n)=n^{-k}f^2(-1/n) $$ Impossible, since $f(1/n)\to a\ne 0$, while $n^{-k}f^2(-1/n)\to 0$, as $n\to\infty$.

B. If $g(0)=1$, then $g(z)=1+z^kf(z)$, where $f$ analytic in the unit disk and $f(0)=a\ne 0$, in which case $$ 1+n^{-k}f(1/n)=g(1/n)=g^2(-1/z) =\big(1+(-n)^{-k}f(-1/n)\big)^2=\\=1+2(-n)^{-k}f(-1/n)+n^{-2k}f^2(-1/n) $$ and hence $$ f(1/n)=2(-1)^kf(-1/n)+n^{-k}f^2(-1/n) $$ which, as $n\to\infty$, tends to $$ a=2(-1)^ka. $$ Contradiction.

Hence, only a constant $g\equiv 0$ or $g\equiv 1$ could satisfy $g(1/n)=g^2(-1/n)$, for all $n>1$.

Iterating $g(z)=g^{2}(-z)$ we get $g(z)=g^{2^{n}}((-1)^{n} z)$. If $|g(z)| >1$ we get $|g(z)|=|g^{2^{n}}((-1)^{n} z)| \to \infty$ as $n \to \infty$ through even values , a contradiction. Hence $|g(z)| \leq 1$ for all $z$ Similarly, $|g(z)| <1$ would imply $g(z)=0$. If $|g(z)|=1$ for all $z$ then $g$ is a constant by MMP. Hence there is no non-constant solution.