= Solution
A <transformation model> consists of a <group> $G$ acting on the <sample space> and on the set of <statistical parameters>, with the compatibility property
$$
P_{g\theta}=g_*P_\theta,\qquad g\in G.
$$
In words, transforming a sample from $P_\theta$ produces a sample from $P_{g\theta}$. A transitive <transformation model> is generated from one reference distribution by the <group action>. For densities, the compatibility property includes the <Jacobian determinant> of the sample transformation; it is not merely equality of <density> values at transformed points.
A <maximal invariant> $A$ is constant on <orbits of a group action> and separates them: $A(gx)=A(x)$, and $A(x)=A(y)$ implies $y=gx$ for some $g$. An <equivariant estimator> $T$ of the parameter satisfies $T(gx)=gT(x)$. Equivariance means the estimate follows the transformed parameter, whereas invariance means the reported value stays unchanged.
First suppose the set of <statistical parameters> can be identified with the <group>, as happens for a simply transitive action. An <equivariant estimator> then provides a group-valued fit $\widehat g(x)$ with $\widehat g(gx)=g\widehat g(x)$. Normalize the sample by this fit:
$$
\boxed{A(x)=\widehat g(x)^{-1}x.}
$$
Indeed $A(gx)=\widehat g(x)^{-1}g^{-1}gx=A(x)$. Conversely, if $A(x)=A(y)$, then $y=\widehat g(y)\widehat g(x)^{-1}x$, placing $x,y$ in the same <orbit of a group action>. This establishes maximality, rather than only invariance.
For a transitive action with a nontrivial <stabilizer> $H$ of a reference parameter $\theta_0$, a parameter-valued <equivariant estimator> does not specify a unique normalizing transformation. Choose a section $h_\theta$ with $h_\theta\theta_0=\theta$ and form $b(x)=h_{T(x)}^{-1}x$. Under $g$,
$$
b(gx)=h_{gT(x)}^{-1}g h_{T(x)}b(x),\qquad h_{gT(x)}^{-1}g h_{T(x)}\in H.
$$
Thus the $H$-orbit of $b(x)$ is invariant. It is maximal: if $b(y)=h b(x)$ with $h\in H$, then $y=h_{T(y)}h h_{T(x)}^{-1}x$. Appropriate <measurable> sections and <orbit of a group action> coordinates are assumed when treating this as a <statistic>. This qualification explains why an arbitrary equivariant <statistic> alone is not automatically a complete set of invariant coordinates.
For the <location-scale family>, write $X_i=\mu+\sigma Z_i$ with $\sigma>0$ and a fixed joint distribution of $Z$. The <group> consists of maps $x_i\mapsto a+bx_i$ with $b>0$, acting on parameters by $(\mu,\sigma)\mapsto(a+b\mu,b\sigma)$. For a nonconstant sample of size $n\ge2$, take the <equivariant estimator>
$$
\widehat\mu=\overline X,\qquad \widehat\sigma=s=\left\{\frac1n\sum_i(X_i-\overline X)^2\right\}^{1/2}.
$$
It gives the <maximal invariant>
$$
\boxed{A(X)=\left(\frac{X_1-\overline X}s,\ldots,\frac{X_n-\overline X}s\right).}
$$
Its coordinates satisfy $\sum_iA_i=0$ and $\sum_iA_i^2=n$. They stay unchanged under location and positive-scale transformations. If two samples have the same standardized <vector>, then
$$
y_i=\overline y+\frac{s_y}{s_x}(x_i-\overline x),
$$
so a single <group> transformation maps the first sample onto the second. The <vector> retains the observation labels: sorting it would discard information not removed by this location-scale <group>. The construction is algebraic and does not require the underlying distribution to have finite <moments>. Constant samples can be assigned a separate invariant value; they form a single exceptional <orbit of a group action> and have probability zero under an independent continuous base distribution.
Finally, in a transitive <transformation model> every invariant <statistic> is ancillary. Choose $g$ taking $\theta_0$ to $\theta$ and use $X_\theta\overset d=gX_{\theta_0}$; invariance gives $A(X_\theta)\overset d=A(X_{\theta_0})$. Thus the standardized residual configuration provides a natural conditioning variable, while the equivariant fit contains the location and scale information.
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