$W$ is a bounded connected open subset of $\mathbb C$. If $\varphi : W \to W $ is an analytic function, such that $\varphi(w) = w$ and $\varphi'(w) = 1$, for some $w \in W$, then $\,\varphi={\rm Id}$.
I'm thinking I need to somehow implement the Schwarz Lemma? But I don't know exactly how to do that here...
First solution. Without loss of generality, assume that $w=0$, and let $r>0$, such that $\overline{D}(0,r)\subset W.\,$ In $f$ is not the identity, then in $\overline{D}(0,r)$ can be expressed as $$ f(z)=z+a_kz^k+\mathcal {O}(z^{k+1}), $$ where $a_kz^k$ is the lowest non-zero term, of order higher than one. Note that the functions $\,f_n=f\circ\cdots\circ f$ ($n$ times), are also analytic on $W$ and $\,f_n[W]\subset W$, for all $n\in\mathbb N$. Also, it is not hard to see that $$ f_n(z)=z+na_kz^k+{\mathcal O}(z^{k+1}), \quad\text{for all $n\in\mathbb N$.} $$ Cauchy integral formula provides that $$ na_k=\frac{1}{2\pi i}\int_{|z|=r}\frac{f_n(\zeta)\,d\zeta}{\zeta^{k+1}}. \tag{1} $$ If $M=\sup_{z\in W}|z|$, then $(1)$ implies that $$ n|a_k|\le \frac{\sup_{z\in\ W}|\,f(z)|}{r^k}=\frac{M}{r^k}, $$ and as this is valid for all $n\ge 1$, then $a_k=0$. Contradiction. Hence $f(z)=z$.
Second solution. Here we use the additional assumption that $W$ is a simply connected region, in which case, $W$ does not even have to bounded. It suffices to be different from $\mathbb C$.
We shall utilise the following results:
Riemann Mapping Theorem. If $\,W \subsetneq\mathbb C$ is simply connected and $\,w\in W,\,$ then there exists a conformal mapping $\,f : W\to\mathbb D$, where $\mathbb D$ is the open unit disk, such that $\,f(w)=0$.
Lemma. If $g:\mathbb D\to\mathbb D$ is analytic, $\,g(0)=0$ and $\,g'(0)=1,\,$ then $\,g(z)=z,\,\,$ for all $z\in\mathbb D$.
Assume now that $W\subsetneq\mathbb C$ is a simply connected region, $\,\varphi:W\to W,\,$ analytic, and there exists a $w\in W$, such that $\varphi(w)=w$ and $\,\varphi'(w)=1$. Riemann Mapping Theorem provides an $\,f:W\to\mathbb D$, with $\,f(w)=0$. Let $\,g=f\circ\varphi\circ f^{-1}$. Clearly, $g$ is analytic in $\mathbb D$, $\,g[\mathbb D]\subset \mathbb D$, and $g(0)=0$ and $$ g'(0)=f'\big((\varphi\circ f^{-1})(0)\big)\varphi'\big((\, f^{-1})(0)\big)(\,f^{-1})'(0)=f'(w)\,\varphi'(w)\frac{1}{f'\big(\,f^{-1}(0)\big)}\\=f'(w)\,\varphi'(w)\frac{1}{f'(w)}=1. $$ Using the Lemma, we obtain that $\,g(z)=z,\,$ and hence $\varphi=f^{-1}\circ g\circ f=\mathrm{Identity}.$