Trivial action on the standard meridians fixes the punctures and the stem arcs up to homotopy

Fixing the punctures. Let $h(q_i)=q_{\sigma (i)}$. A positive small circle about $q_i$ is carried to a positive Jordan circle about $q_{\sigma (i)}$ containing no other marked point. Its lasso represents a conjugate of $x_{\sigma (i)}$: contract the circle inside its once-punctured neighborhood to a small circle and compare its tether with the standard tether. Abelianization in the free basis sends this conjugate to $e_{\sigma (i)}$, whereas the hypothesis $h_{\ast }{x}_i=x_i$ sends it to $e_i$. Hence $\sigma (i)=i$ for each $i$.

Compactifying the tether calculation correctly. Fix $i$ and abbreviate $q=q_i$, $s=s_i$, $a=h\circ s$. Take a small round disk $B$ about $q$ avoiding every other marked point. Continuity of $a$ at its endpoint gives a terminal segment contained in $B$. In $B\setminus \{q\}$ write that terminal segment as $(r(u),\theta (u))$ using a continuous lift of its polar angle on the parameter interval. Replace it, relative to its initial point and $q$, by the radial segment: interpolate its angle to the initial angle and its positive radius to the linear radius of that segment. For parameters below $1$ all radii remain positive; at $1$ the radii tend uniformly to zero during the interpolation, since both original and linear radii do so. Thus this is a homotopy on the compact square, avoiding $q$ except at the endpoint, even when $\theta (u)$ is unbounded. Adjust the terminal angle and a connecting path along a circle to obtain a representative consisting of a path $P:d\to p$ followed by the fixed radial tail $p\to q$ of $s$, for a point $p$ on a sufficiently small circle $C\subset B$. Denote the truncated standard stem $d\to p$ by $S$. The connecting-circle adjustment has the same compact homotopy description.

Equality of meridians controls the tether. The lasso associated with $a$ represents $h_{\ast }{x}_i=x_i$. In the terminal modification of step 1.2 the small circles are positive generators of ${\pi }_1(B\setminus \{q\})$; changing the terminal tether conjugates that generator within this cyclic group and leaves it unchanged. Consequently $[PC{P}^{-1}]=[SC{S}^{-1}]=x_i$. Put $w=[P{S}^{-1}]\in {\pi }_1(X,d)$. Then $w{x}_i{w}^{-1}=x_i$. In the free basis this forces $w=x_i^m$ for an integer $m$: in a reduced word write $w=x_i^{a}v{x}_i^b$, where $v$ is empty or its first and last letters are neither $x_i$ nor $x_i^{-1}$. If $v$ is nonempty, the subword $v{x}_i{v}^{-1}$ is reduced and retains a letter other than $x_i^{\pm 1}$, even after adjoining the outer powers. It therefore cannot reduce to $x_i$. Hence $v$ is empty and $w$ is a power of $x_i$.

A peripheral power disappears at a marked endpoint. By [F2], $w=x_i^m$ implies a homotopy of paths with fixed endpoints in $X$ from $P$ to $S{C}^m$ (append $S$, then cancel the backtracking path). Attach the same radial tail $p\to q$ to this homotopy; its compact image in $X$ stays away from the finite set $Q_n$, and its unchanged tail supplies a continuous extension at $q$, uniformly in the homotopy parameter. Finally $C^m$ followed by that tail is homotopic to the tail inside $B$, with $p,q$ fixed: lift its polar angle along its parameter, interpolate it to the constant angle, and interpolate the radius to the positive linear radius ending at zero. The resulting paths avoid $q$ in their interiors; uniform convergence of their radii to zero again proves continuity on the compact square. Thus $a\simeq s$ in the relative-endpoint sense asserted. This concerns a peripheral power at the endpoint, and does not contract a nontrivial meridian loop inside $X$.

Conclusion. Step 1.1 proves that every puncture is fixed, and step 3.1 supplies the asserted compact relative-endpoint homotopy for each stem. The radii, paths and homotopies involve finitely many given arcs and explicit polar interpolations; no infinite selection or choice axiom is used. The intermediate paths need not be embeddings; upgrading this homotopy to an isotopy is a separate proper-arc result.