Why rotating a function in this paper uses $R^{-1}$ and not $R$?

Question

If you look at Eq. 1 of this (paper)[http://arxiv.org/abs/1801.10130], it states that

"we introduce the rotation operator $L_R$ that takes a function $f$ and produces a rotated function $L_Rf$ by composing $f$ with the rotation $R^{−1}$: $$ [L_Rf](x) = f(R^{−1}x). $$ Due to the inverse on $R$, we have $L_{RR′} = L_RL_R′$."

1. It seems something trivial, but I cant get my head around it that if $L_R$ rotates the function, then should not $[L_Rf](x) = f(Rx)$, i.e. rotated function being equivalent to function on rotated input? (as main thrust of paper is towards equivariance.)

2. Why does $R^{-1}$ help in meeting the condition $L_{RR′} = L_RL_R′$ and not $R$ itself?

Answer 1

$$L_RL_{R'}(f)(x)=L_Rf(R'^{-1}x)$$$$=f(R'^{-1}R^{-1}x)=f((RR')^{-1}x)$$$$=L_{RR'}(f)(x)$$ The important fact is that $(RR')^{-1}=R'^{-1}R^{-1}$.