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Let $G$ be a Lie group acting on a manifold $M$. Then for any $g\in G$ we can define the map $\phi_g\colon M\to M$ by $m\mapsto g\cdot m$. This gives rise to a representation of $G$ on the ring of differential forms $\Omega(M)$ via $$\rho_g\omega=(\phi_g^{-1})^*(\omega).$$ The Lie algebra also has the respresentation $$L_X\omega= \frac{d}{dt} (\rho_{\exp t X} \omega)|_{t=0}.$$

With these definitions, I can easily show that $\rho_g L_X \rho_g^{-1}=L_{Ad_g X}$. I am reading a text where it is claimed that $$\rho_g \iota_X \rho_g^{-1}=\iota_{Ad_g X}.$$

(1) I'm not quite sure about the meaning of the interior product in this context. The only reasonable one seems to be the contraction by $L_X$ since $L_X$ can be applied to $0$-forms, and we can think of it as a vector field on $M$.

(2) Can someone please sketch the proof of the claim please? (The reason that I find it confusing it that if $\omega$ is an $n$-form, then $\iota_X \rho_g^{-1}$ is an $(n-1)$-form and when we apply $\rho_g$ to this $(n-1)$ form, according to my interpretation, $\rho_g$ does not affect the first slot of $\omega$. It is not quite clear to me how contraction with $Ad_g X$ is created on the right because this concerns exactly the first slot of $\omega$.)

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