= Solution
A <transformation model> consists of a <group> $G$ acting measurably and invertibly on the sample space, an induced <group action> on the parameter space, and a family $\{P_\theta\}$ satisfying $Y\sim P_\theta\Rightarrow gY\sim P_{g\theta}$. In a generated, transitive model, every $P_\theta$ is the image of a reference distribution under some transformation. In the density formulation, the transformed density includes the appropriate sample-space <Jacobian determinant>; the model is a statement about whole <probability laws>, not merely about changing parameter names.
A <maximal invariant> $A$ has two properties: $A(gy)=A(y)$ for every $g$, and $A(y)=A(z)$ implies $z=gy$ for some $g$. It is therefore a coordinate for the <group orbits>, retaining exactly the information unaffected by the transformations. An <equivariant estimator> obeys $T(gy)=gT(y)$, where the right action is the induced parameter action.
For the normalization construction, suppose the estimator determines a <group> element $t(y)\in G$ with $t(gy)=gt(y)$. Define $A(y)=t(y)^{-1}y$. Then
$$
A(gy)=(gt(y))^{-1}gy=t(y)^{-1}y=A(y).
$$
If $A(y)=A(z)$, rearranging gives $z=t(z)t(y)^{-1}y$. This is an actual transformation in $G$, so \b[the normalized sample is a <maximal invariant>]. This <group>-valued assumption matters: an estimator of a parameter with a nontrivial stabilizer does not necessarily determine a unique normalization. For a transitive parameter space $G/H$, choose a transformation carrying the estimated parameter to the reference value and take the resulting normalized sample modulo the remaining $H$-action. Different choices then have the same residual orbit. Under a free action $H$ is trivial and the simpler formula applies.
For a <location-scale model>, $g_{a,b}(y_1,\ldots,y_n)=(a+by_1,\ldots,a+by_n)$, $b>0$, with composition $(a,b)(c,d)=(a+bc,bd)$. Its parameter action is $(\mu,\sigma)\mapsto(a+b\mu,b\sigma)$. Choose location and scale estimates satisfying
$$
\widehat m(a+by)=a+b\widehat m(y),\qquad \widehat s(a+by)=b\widehat s(y)>0.
$$
For instance, on nonconstant samples take the <sample mean> and the positive square root of $n^{-1}\sum_i(y_i-\bar y)^2$. The standardized vector
$$
\boxed{A(y)=\left(\frac{y_i-\widehat m(y)}{\widehat s(y)}\right)_{i=1}^n}
$$
is invariant. Conversely, equality of these vectors gives $z_i=c+dy_i$ for all $i$, with $d=\widehat s(z)/\widehat s(y)>0$ and $c=\widehat m(z)-d\widehat m(y)$. Hence it is maximal, not just invariant. Degenerate constant samples have zero fitted scale and form a separate orbit; define a separate invariant value there. The construction is on labelled sample vectors: one must not discard their order unless permutations are also included in the transformation <group>.
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