I have a somewhat stupid question whose answer will most likely be "Obviously, yes." I just want to make sure I am getting the technical details down before I move forward.
Let $\psi_0 = (X, \Phi, X^{\vee},\Phi^{\vee})$ be a root datum, where $X= X(T)$. To any automorphism $f$ of $X$, let $^t f$ denote the dual isomorphism of $X^{\vee}$. Then $f \mapsto \, ^t f$ is a bijective antihomomorphism.
An automorphism of root data is an isomorphism $X \rightarrow X$ which permutes $\Phi$, and whose dual map $X^{\vee} \rightarrow X^{\vee}$ permutes $\Phi^{\vee}$.
Let $\psi_0^{\vee} = (X^{\vee}, \Delta^{\vee}, X, \Delta)$ be the dual root datum. Then $f \mapsto \, ^t f$ gives a bijective antihomomorphism $\textrm{Aut}(\psi_0) \rightarrow \textrm{Aut}(\psi_0^{\vee})$.
One has an interest in making an isomorphism between these last two groups, rather than a bijective antihomomorphism. For example, in the definition of L-group (Borel, Automorphic L-functions in Corvallis II), $\psi_0$ is the based root datum of a connected, reductive $k$-group $G$, and $\psi_0^{\vee}$ is the root datum of a $\overline{k}$-group $G^{\vee}$, and one composes various maps to obtain a homomorphism
$$\textrm{Gal}(k_s/k) \rightarrow \textrm{Aut}(\psi_0) \rightarrow \textrm{Aut}(\psi_0^{\vee}) \rightarrow \textrm{Aut}(G^{\vee})$$
and the $L$-group $^L G$ is defined as the semidirect product of $G^{\vee}$ with $\textrm{Gal}(k_s/k)$. In order for this composition to be a homomorphism, the middle map should be defined by the formula
$$f \mapsto \, ^t f^{-1}$$
rather than $f \mapsto \, ^t f$. Is this correct?