Let $G$ be a reductive (or just semisimple) algebraic group over an algebraically closed field $k$ of characteristic $p > 0$, let $T$ be a maximal Torus and let $f:G \rightarrow G$ be an isogeny. Suppose the induced morphism of root data $f^\star: \Psi(G,T) \rightarrow \Psi(G,T)$ is a Frobenius morphism multiplying roots by $q = p^r$ (confer 9.6.3 in Springer's book on algebraic groups). Is it true that $f$ is the geometric Frobenius of an $\mathbb{F}\_{q}$-structure on $G$?


Equivalent formulation: Is the morphism $F:G \rightarrow G$ defined by $F( u\_\alpha(c) ) = u\_\alpha(c^q)$ for all roots $\alpha$ and $F(t) = t^q$ for all $t \in T$ the geometric Frobenius of an $\mathbb{F}\_{q}$-structure on $G$?


I'm pretty sure this is true (perhaps under some additional conditions). I know that there is an easy characterization of Frobenius morphisms in terms of the comorphism of the coordinate rings but I was not able to verify this.



If the morphism of root data induced by an isogeny of a reductive group is a Frobenius, is then the isogeny itself a Frobenius?
The answer is "yes", but not in a good way: the descent it arises from is the split form, so this does not encode an interesting $\mathbf{F}\_ q$-structure. More precisely, $f$ arises from the $q$-Frobenius of a split descent down to $\mathbf{F}\_ q$ (and so even to the prime field $\mathbf{F}\_ p$). In particular, this is definitely not the way to encode the information of "interesting" descents of $(G,T)$ to $\mathbf{F}\_ q$, if that may be your ultimate intention. 


To see this, let $(G\_0, T\_0)$ be the split form over $\mathbf{F}\_ q$, and $F\_0:(G\_0, T\_0) \rightarrow (G\_0,T\_0)$ the $q$-Frobenius morphism. This induces the self-map of the root datum given by $q$-multiplication on ${\rm{X}}(T\_0)$. Extending scalars to $k$ has no effect on the root datum for split groups, so $(F\_ 0)\_ k$ induces the endomorphism of the root datum of $((G\_ 0)\_ k, (T\_ 0)\_ k)$ given by $q$-multiplication on the root datum. Pick a $k$-isomorphism between the $k$-split pairs $(G,T)$ and $((G\_ 0)\_ k, (T\_ 0)\_ k)$ (as we may, by the 
isomorphism Theorem). The resulting isomorphism of root data identifies the maps from $(F\_ 0)\_ k$ and $f$ since $q$-multiplication is functorial with respect to all homomorphisms between abelian groups. Thus, these maps coincide up to the action of some $t \in \overline{T}(k) = (\overline{T}\_ 0)(k)$, where 
$\overline{T} := T/Z\_G$ and 
$\overline{T}\_ 0 := T\_ 0/Z\_ {G\_ 0}$ are the "adjoint tori''. Using an isomorphism 
$\overline{T}\_ 0 \simeq \mathbf{G}\_m^r$, $t$ goes over to some $r$-tuple $(t\_i)$ with $t\_i \in k^{\times}$. Since $k$ is separably closed, we can solve $t\_i = y\_i^{q-1}$ with $y\_i \in k^{\times}$. In other words, $t = f(y)/y$ for some $y \in T\_0(k)$. So if we modify the identification of $(G\_0, T\_0)$ as an $\mathbf{F}\_ q$-descent of $(G,T)$ by composing with the action of $y$ or $y^{-1}$ (depending on direction of maps) then $f = (F\_ 0)\_ k$ via the new identification of $(G\_ 0, T\_ 0)$ as a split $\mathbf{F} \_q$-descent of $(G,T)$.
QED


Remark: A refinement of the argument (using some care if $k$ is not algebraic over 
$\mathbf{F}\_ p$) proves that the split descent to $\mathbf{F} \_q$ is the only one which can work, but I won't get into that here. (It comes down to the fact that the solutions to $y^{q-1} = 1$ in $k^{\times}$ lie in $\mathbf{F} \_q$.) 


Note that the above works over any separably closed field of characteristic $p > 0$, not necessarily algebraically closed, provided the Isomorphism Theorem is valid over such fields. The Existence, Isomorphism, and even Isogeny Theorems are valid for split connected reductive groups over any field. The paper of Steinberg mentioned in the previous answer is really nice, but unfortunately assumes from the outset that the ground field is algebraically closed. Good news, in case you may care, is that one can deduce the results over any field from that case as a "black box'' by using descent theory. See Appendix A.4 of the book "Pseudo-reductive groups''. 


