Consider a group $\,G\,$, a vector space $\,{\mathbb{V}}\,$, and a space $\,{\mathbb{W}}\,$ of maps $\,\varphi:\,G\longrightarrow{\mathbb{V}}\,$. In the space $\,{\mathbb{W}}\,$, acts a natural representation
$$
U_g\varphi(x)=\varphi({g^{-1}}x)~~,\qquad g,\,x\in G~~.
$$
It is easy to prove (see below) that an arbitrary representation $\,A(G)\,$ on $\,{\mathbb{V}}\,$,
$$
A~:\quad G~\longrightarrow~GL({\mathbb{V}})~~,
$$
is equivalent to the representation $U$ on ${\mathbb{W}}$. This, however, seems to entail a paradoxical result that all representations on ${\mathbb{V}}$ are equivalent. (Or does it not?) I must be missing something here.
Now, the Proof.
Introduce an operator $$ {\mathfrak{M}}_{{_{\rm{A}}}}\,:\quad{\mathbb{V}}\,\longrightarrow\,{\mathbb{W}} $$ mapping an element $\,v\in{\mathbb{V}}\,$ to a function $\,\varphi\in{\mathbb{W}}\,$, $$ \varphi(x)~=~( {\mathfrak{M}}_{{_{\rm{A}}}}\,v)\,(x)~~, $$ where $\,x\in G\,$ is a free variable. We set the mapping as $$ \varphi(1)\,=\,v~~,\quad\varphi(x)\,=\,A(x^{-1})\,v~~.~~~~~~~~~~~~~~~~~~(1) $$ The operator $\,{\mathfrak{M}}_{{_{\rm{A}}}}\,$ is invertible, because
$ \mbox{(a)} \;\;\forall\varphi\in{\mathbb{W}}\,, ~\exists~{\mathfrak{M}}_{{_{\rm{A}}}}^{-1}:~\varphi\,\longrightarrow\,\varphi(1)~,~~ \mbox{so} ~~ {\mathfrak{M}}_{{_{\rm{A}}}}~ \mbox{is surjective}, $
$ \mbox{(b)} \;\;\operatorname{Ker}\, {\mathfrak{M}}_{{_{\rm{A}}}}\,=\, 0\,, ~~\mbox{so} ~~{\mathfrak{M}}_{{_{\rm{A}}}}~ \mbox{is injective}. $
(For $\,v\neq\vec{\,0}\,$ and $\,\operatorname{det} A(x^{-1})\neq 0\,$, we always get a nonzero $\,A(x^{-1})\, v\,$.)
Similarly, a vector $\,A(g)v\,$, for a fixed $\, g\in G\,$, will be mapped by $\, {\mathfrak{M}}_{{_{\rm{A}}}}\,$ to a function $$ \varphi^{\,\prime}(x)\,=\,( {\mathfrak{M}}_{{_{\rm{A}}}}\, A(g)\, v\,)\,(x) $$ such that $$ \varphi^{\,\prime}(1)\,=\,A(g)v\,~,\quad\varphi^{\,\prime}(x)\,=\,A(x^{-1})\,A(g)\,v\,~.~~~~~~~~~~~~~~(2) $$ We now can write $$ U_g\,( {\mathfrak{M}}_{{_{\rm{A}}}}\,v)\,(x)\,=\,U_g\,\varphi(x)~=~\varphi (g^{-1}x)~=~A(\,(g^{-1}x)^{-1}\,)\,v $$ where, at the last step, we used definition (1) with $\,g^{-1}x\,$ used instead of $\,x\,$. With aid of (2), the above can be continued as $$ =\,A(x^{-1})\,A(g)\,v~=~\varphi^{\,\prime}(g)\,=\,( {\mathfrak{M}}_{{_{\rm{A}}}}\, A(g)\, v\,)\,(x)~~, $$ so $\, {\mathfrak{M}}_{{_{\rm{A}}}}\,$ is an intertwiner: $$ U\, {\mathfrak{M}}_{{_{\rm{A}}}}\,=\, {\mathfrak{M}}_{{_{\rm{A}}}}\, A~~. $$ Moreover, since $\, {\mathfrak{M}}_{{_{\rm{A}}}}\,$ is invertible, the representations $\,A\,$ and $\,U\,$ are equivalent. Thus we have proven a somewhat counterintuitive Theorem:
Any representation $\, A(G)\,$ on the elements $\,v\,$ of a space $\,{\mathbb{V}}\,$ is equivalent to the natural representation $\, U_g\varphi(x)=\varphi(g^{-1}x)\,$ on the maps $\,\varphi:\, G\longrightarrow{\mathbb{V}}\,$.
This result looks suspicious.
Indeed, were an arbitrary representation $\,A(G)\,$ on $\,{\mathbb{V}}\,$ equivalent to $\,U(G)\,$ on $\,{\mathbb{W}}\,$ of maps, would this not ensure the mutual equivalence of all representations on $\,{\mathbb{V}}\,$? That would be hard to believe.
Where have I made a mistake?