Is the converse of this statement true? we know that:

Ratios of scale functionals are location and scale invariant.

My question is: is it true that every scale and location 
invariant functional can be rewritten as the ratio of two scale
 and location equivariant functional? 
It seems logical to me, but I can't find a source or a proof 
 for this. 

Edit: 
to respond to whuber's question below. Here I mean functional 
in the statistical sense; e.g a map from a set of probability 
distributions to the real numbers. 
Examples include the mean functional $\mu$:
$$\mu(F)=\int_{-\infty}^{\infty}xdF(x),$$
on the set of distributions with finite first moments or the variance functional $\sigma^2$:
$$\sigma^2(F)=\int_{-\infty}^{\infty}(x-\mu(F))^2dF(x).$$
A non-negative functional $T$ is a scale functional if 
$$T(G)=|\lambda|T(F)$$ 
when $G(x)=F((x-\mu)/\lambda)$ and $\lambda\neq0$ (i.e. if $T$ is 
scale equivariant and shift invariant).
A functional $T$ is a scale and shift invariant if
$$T(G)=T(F)$$ 
when $G(x)=F((x-\mu)/\lambda)$ and $\lambda\neq0$.
 A: Here is my understanding of the terminology.  $\mathcal{X}$ is the set of all distributions on the real line.  For $F\in\mathcal{X}$, $\mu\in\mathbb{R}$,and $\sigma\in\mathbb{R}-{0}$, define a transformation from $F\to F$ via
$$(T_{\mu,\sigma}(F))(x) = F((x-\mu)/\sigma)(x)$$
for all $x\in \mathbb R$.  (This is the action of the affine group of the real line induced on the set of measures on the real line.)  A functional S is a map $S:\mathcal{X}\to \mathbb{R}$.  It is invariant when
$$S[T_{\mu,\sigma}(F)] = S[F]$$
for all $F$, $\mu$, and $\sigma$, and it is equivariant when
$$S[T_{\mu,\sigma}(F)] = |\sigma|S[F]$$
for all nondegenerate $F$, $\mu$, and $\sigma$.

The idea is to find a natural location and scale for all relevant distributions.  Here is one way.
For all $F$ and $0\lt q \lt 1$, the set $$\{x\in \mathbb{R}\,|\, F(x)\ge q\}$$ is nonempty and must have a lower bound, whence it has a greatest lower bound $F_{[q]}$.  Define $$m(F) = F_{[1/2]}.$$  The set $$\{x\in \mathbb{R}_{+}\,|\, F(m(F)+x) - F(m(F)-x)\ge q\}$$ is bounded below by $0$ and nonempty, whence it must have a greatest lower bound $F^\prime_{[q]}$.  Moreover, for $q$ sufficiently large, this glb must be strictly positive provided $F$ is nondegenerate.  The set of such $q$ for which $F^\prime_{[q]} \gt 0$ has a greatest lower bound $q^\prime_F$.  Define  $$s(F) = F^\prime_{[(1-q)/2]}.$$
It follows that $m(F)$ (the location) and $s(F)$ (the scale) are well-defined and $s(F)\gt 0$.
It is straightforward to check that
$$m(T_{\mu,\sigma}(F)) = m(F) + \mu$$
and
$$s(T_{\mu,\sigma}(F)) = s(F)|\sigma|$$
for $\sigma\ne 0$ (which makes $s$ an equivariant functional).  Choosing $\mu=-m(F)$ and $\sigma = 1/s(F)$ yields a transformation
$$Z: F \to F^{0} = T_{-m(F),1/s(F)} (F)$$
which is well-defined on all non-degenerate distributions.  $F^{0}$ is the standardized version of $F$ and
$$F = T_{m(F),s(F)}(F^{0}).$$
In other words, every nondegenerate distribution is a shifted and scaled version of its standardized version.

Here is one solution.  Let $S$ be any invariant functional.  It is immediate that
$$S_1[F] = s(F)S[F^{0}]$$
defines an equivariant functional, exhibiting
$$S = \frac{S_1}{s}$$
explicitly as the ratio of two equivariant functionals, because $S[F] = S[F^{0}]$.
