This question concerns Springer's "Linear Algebraic Groups" 2nd Edition page 69.
Let $G$ be a linear algebraic group. For $x\in G$ we have an automorphism $Int(x):G \to G$, $y \mapsto xyx^{-1}$ fixing $e$ and therefore inducing linear automorphisms $$Ad \,x:T_eG \to T_eG,$$ $$(Ad\, x)^*:(T_eG)^* \to (T_eG)^*$$ of the tangent and cotangent space. Thus for $u\in (T_eG)^*$ we have $$(Ad \,x)^*(u)(X) = u(Ad(x^{-1})X) \quad (x\in G,\, X\in T_eG).$$
Why is the right side of the equation not $u(Ad(x)X)$?
Dont we have the identity $\langle f^*(\varphi),v\rangle = \langle \varphi, f(v)\rangle $ for $\langle\_\, ,\,\_\rangle $ the natural pairing of dual spaces, $v\in V$, $ \varphi \in V^*$, $f:V\to V$ and therefore
$$\begin{align} (Ad \,x)^*(u)(X)&=\langle (Ad\, x)^*(u),X\rangle \\ &= \langle u, (Ad\, x)(X)\rangle \\ &=u((Ad\, x)(X))? \end{align}$$