Schwarz's Lemma, fixed points question

This is from an old qualifying examination question.

If f is holomorphic in the unit disk $D$ and $|f(z)|<1$ for all $z\in D$. Suppose also that $f$ has two distinct fixed points in $D$ then $f(z)=z$ for every $z\in D$

I know that I have to use the Schwarz's lemma and may be make use of Möbius transformations. I tried setting $g(z)=\phi_a\circ f\circ \phi_{-a}$. But that does not seem to work because I don't see why $|g(z)|=|z|$ for some non zero $z$.

Any helpful hints are greatly appreciated.

Edit: Of course the $a$ above is one of the fixed points.

Hint: Let $b$ be the other fixed point of $f$. What happens when you apply $g$ to $\phi_a(b)$?

Here's a full solution. Let $a$ and $b$ be the distinct fixed points of $f$ and let $\varphi$ be the conformal automorphism of the disk sending $a$ to $0$, recall

$$\varphi(z)=\frac{z-a}{1-\bar{a}z}.$$

In particular it's not hard to check that $g=\varphi\circ f \circ \varphi^{-1}$ fixes $0$. We also have

$$ g(\varphi(b))=\varphi[f(\varphi^{-1}[\varphi(b)])]=\varphi(f(b))=\varphi(b)$$

by Schwar'z lemma $g(z)=cz$. Now $\varphi(b)\neq 0$ since $\varphi(a)=0$ thereby $g(z)=z$, in particular this gives that $\phi^{-1}=f \circ \phi^{-1}$ so $f$ must be the identity on the disk.