Notation
I'm working through Bröckner and Dieck's Representations of Compact Lie Groups. On page 18 they define the Adjoint map $$ \mathrm{Ad}: G \to \mathrm{Aut}(LG), \quad g \mapsto \mathrm{L}c(g):= T_e c(g), $$ where $ G $ is a Lie group, $ LG $ is its Lie Algebra, $T_e$ is the pushforward operator, and $$ c(g): G \to G, \quad x \mapsto g x g^{-1}$$ is the inner automorphism. They also denote the integral curve of the vector field $Z \in LG$, which starts at $e$, by $\alpha^Z$.
Question
My question relates to the following identity which they do not seem to fully justify: $$ \mathrm{Ad} (\alpha^X (s))Y = \frac{\partial}{\partial t} \Big|_0 c(\alpha^X (s)) \alpha^Y (t). $$ At first I was not able to derive the identity, but I had a bit of insight after having typed out a bit of my workings. I believe I've proven it below. I'm fairly new to this notation so I would like someone to let me know if there are any errors.
Work so far
$$ \begin{aligned} \mathrm{LHS} &= \mathrm{Ad} (\alpha^X (s)) Y \\ & = Lc (\alpha^X (s)) Y = T_e c (\alpha^X (s)) Y = Y \circ {c(\alpha^X (s))}^* \\ \implies \mathrm{LHS}(\phi) &= (\mathrm{Ad} (\alpha^X (s)) Y)(\phi) = Y \circ {c(\alpha^X (s))}^* \phi = Y(\phi \circ c(\alpha^X(s))) \\ &= \Big( \frac{\partial}{\partial t} \Big|_0 \alpha^Y (t)\Big) (\phi \circ c(\alpha^X(s))) = \frac{\partial}{\partial t} \Big|_0 \phi (c(\alpha^X (s)) \alpha^Y (t)) = \mathrm{RHS}(\phi). \end{aligned} $$
Remember that $(\alpha^Y)'(0) = Y$ and by definition of conjugation $$c(\alpha^X(s))\alpha^Y(t) = \alpha^X(s) \alpha ^Y(t) \alpha^X(-s) $$
just remember the property $\alpha^X(s)^{-1} = \alpha^X (-s)$. Now the adjoint representation is the left invariant vector field whose value at $e \in G$ is $$ \frac{d}{dt}\left(c(\alpha^X(s))\alpha^Y(t)\right)\bigg|_{t=0} = (c(\alpha^X(s))_{\ast e} (\alpha^Y)' (0) = \mathrm{Ad}\ (\alpha^X (s)) Y$$