I ran into this while studying for my qual.
Suppose $S=\{z:-\pi/2<\Im\{z\}<\pi/2\}$ and there is an entire function $g$ with $g(S)\subseteq (S)$. If $g(-1)=0$ and $g(0)=1$ prove that $g(z)=z+1$.
I tried to use Schwarz's lemma but to get from disc to strip or vice-versa the conformal maps involve logs and exps and things get ugly with estimates. Another approach would be to show that $g$ is an analytic automorphism of $\mathbb{C}$ in which case we'd also be done; but, I am not sure how to get there without any injectivity assumptions on $f$.
Let us consider the conformal map $\varphi $ from the unit disc, $\mathbb{D}=\{z\in \mathbb{C} : |z|<1\}$, to the strip, $S=\{z \in \mathbb{C} : |\text{Im } z|<\frac{\pi}{2}\}$, i.e. $\varphi:\mathbb{D}\rightarrow S$ given by \begin{align} \varphi(z) = \text{Log}\left(\frac{1+z}{1-z}\right) \end{align} and the translation map $\tau_{-1}:\mathbb{C}\rightarrow \mathbb{C}$ given by \begin{align} \tau_{-1}(z) = z-1. \end{align} Then we see that \begin{align} h:= \varphi^{-1}\circ\tau_{-1}\circ g\circ\varphi: \mathbb{D}\rightarrow \mathbb{D} \end{align} and $h(0) = (\varphi^{-1}\circ\tau_{-1}\circ g)(0) = \varphi^{-1}(0) = 0$. Moreover, we also see that \begin{align} h\left(\frac{1-e}{1+e}\right) = \frac{1-e}{1+e} \end{align} then by Schwarz's lemma we have that $h(z) = z$. In particular, it follows \begin{align} z = (\varphi^{-1}\circ\tau_{-1}\circ g\circ\varphi)(z) \ \ \ \implies \ \ \ (\tau_{-1}\circ g)(w) = w \ \ \ \implies \ \ \ g(w) = w+1 \end{align} where $w = \varphi(z)$.