Let's say I have a sequence of injective holomorphic maps $f_n \colon \mathbb{D} \to \mathbb{D}$ such that $f_n(0) = 0$. The main thing is that $f$ "almost preserves norms" in the sense that for all nonzero $z$,
$$ r_n |\,z\,| < |\,f(z)\,| < |\,z\,| $$
where $r_n \nearrow 1$. Does it follow that the $f_n$ approximate rotations? More precisely, let's assume that $f_n'(0) > 0$ for every $n$ (in fact by Schwarz lemma-esque reasoning we can get that $f_n'(0) \to 1$). Is it true that $f_n \to \text{Id}$ pointwise/uniformly on compacts?
This was inspired by an exercise in Stein and Shakarchi's complex analysis book.
Edited to add: My motivation in asking this question comes from the Riemann mapping theorem. I'm interested in the relationship between the magnitude of the derivative of a mapping $f \colon K \to D$ at the origin and the size of the largest disk it contains. Indeed such a $K$ must contain a disc about the origin, and if you assume that $|\,f(z)\,| > |\,z\,|$ for all $z \neq 0$ you get that $f$ increases the size of $K$, and also that $|\,f'(0)\,| > 1$. If you keep expanding $K$ by chaining these maps, the "tail expansions" from large domains to $\mathbb{D}$ should look somewhat like rotations, I think. And if they look like rotations the only way for their derivative at the origin to be positive is for them to be almost trivial rotations.
Yes, $f_n(z)\to z$ uniformly on compact subsets of $\mathbb{D}$. Fix such a compact set $K$ and let $r<1$ be such that $K\subset \{z:|z|<r\}$. The image of the circle $|z|=r$ is a simple closed curve that winds around the origin (by the argument principle) and lies in the annulus $r_nr<|z|<r$.
It follows that the domain bounded by this curve converges to the disk $\mathbb{D}_r=\{z:|z|<r\}$ in the kernel sense: see Carathéodory kernel theorem. The theorem implies that $f_n$ converge, uniformly on compact subsets of $\mathbb{D}_r$, to the normalized conformal map of $\mathbb{D}_r$ onto $\mathbb{D}_r$, i.e., the identity map.
Without the assumption that $f_n$ is injective the statement would be false, as is demonstrated by $f_n(z)=zg_n(z)$ with $g_n$ mapping $\mathbb{D}$ onto the cut annulus $\{\frac{n-1}{n}<|z|<1\}\setminus [-1,0]$ so that $g_n(0)>0$.