I have a brain lag. I'm overthinking probably. I have the following equation: $$ f^{-1}(f(x)+f(y))=(g\circ h)^{-1}((g\circ h)(x)+(g\circ h)(y)), $$ where, say, $f,g,h$ are automorphisms on $\mathbb{R}$. Is it enough to compare the relevant parts of both sides of equation to deduce that: $$ f\equiv g\circ h? $$ What it comes from?
EDIT: Ok, the naming is confusing. I mean $f, g, h\colon\mathbb{R}\to\mathbb{R}$ are bijections.
By manipulating this equation and making substitutions with $x$ and $y$, we may arrive at the equation $$(g\circ h\circ f^{-1})(x+y)=(g\circ h\circ f^{-1})(x)+(g\circ h\circ f^{-1})(y)$$ If you add the restriction that $g\circ h\circ f^{-1}$ is continuous (at at least one point), this reduces to Cauchy's functional equation, whose solution allows us to conclude that $$(g\circ h\circ f^{-1})(x)=ax$$ for some nonzero constant $a$, or that $$(g\circ h)(x)=af(x)$$ Which is slightly more general than the conclusion that you proposed.