The mapping class group of the closed unit disk $D_n\subset\mathbb{C}$ with $n$ equally spaced points $P$ on the x-axis (0,1) removed is defined as the set of isotopy classes of self-homeomorphisms $D_n\rightarrow D_n$ which fix the boundary: $\phi\partial D_n=\partial D_n$, and permute the distinguished points: $\phi P=P$.
I am considering a representative $\phi:D_n\rightarrow D_n$ of an element of said mapping class group that induces the identity-automorphism of the fundamental group $\pi_1(D_n,x_0)$, i.e. $\phi_*=\text{id}_{\pi_1(D_n,x_0)}$, where $x_0\in\partial D_n$. I am convinced that $\phi$ has to be homotopic to the identity map $\text{id}_{D_n}$ but am unable to prove it. I know that since $\phi$ is a homeomorphism, $\phi^{-1}_*$ has to be the identity-automorphism as well, but I fail to see how I could use this.
Also, since $\phi$ fixes all loops, for every point $x$ I can choose some loop $\gamma$ on which $x$ lies, and get some homotopy $H_\gamma$ to a loop $\phi\circ\gamma$, on which $\phi(x)$ lies. But I don't see how I could combine those homotopies to get a well defined homotopy from $\text{id}_{D_n}$ to $\phi$, or whether that is even possible at all.
A hint would be appreciated.
Hint: Choose a basepoint-preserving homotopy equivalence $f:X\to D_n$, where $X$ is a wedge of $n$ circles. Since $f$ is a homotopy equivalence, it suffices to show that $\phi\circ f\simeq id_{D_n}\circ f=f$. To show this, you can show that there is a basepoint-preserving homotopy between $\phi\circ f$ and $f$ when restricted to each of the circles of $X$.