Do rational points in a split reductive group act transitively on the orbits of the Cartan subalgebra (w.r.t. automorphism group of Lie algebra)? Let $(G,T,M)$ be a split reductive group (over say, the integers), with Lie algebra $(\mathfrak{g}, \mathfrak{t})$, and let $R$ be a commutative ring.  When $R$ is an algebraically closed field, it is a well-known theorem (of Cartan for $R = \mathbb{C}$, and Chevalley in characteristic zero, and I think Humphreys in general) that all Cartan $R$-subalgebras of $\mathfrak{g}_R$ are $G(R)$-conjugate, but this fails for arbitrary $R$.  For example, when $R$ is the field of real numbers and $G=\operatorname{SL}_2$, the Lie algebra of a non-split maximal torus is not $G(R)$-conjugate to $\mathfrak{t}_R$).  Instead of this general conjugacy, I'd like to know how well the action of $G(R)$ approximates the action of the group of $R$-Lie algebra automorphisms of $\mathfrak{g}_R$.
Question 1: Does $G(R)$ act transitively on the $\operatorname{Aut}_{R\text{-Lie}} \mathfrak{g}_R$-orbit of $\mathfrak{t}_R$?
There is a group scheme analogue of this:
Question 2: Does $G(R)$ act transitively on the $\operatorname{Aut}_{R\text{-gp}.} G_R$-orbit of $T_R$?
It seems pretty clear that a positive answer to the first question implies a positive answer to the second, but I don't know if the reverse holds.  As far as I can tell, the answer to both questions is positive when the canonical isogeny $Z(G) \times D(G) \to G$ is an isomorphism (that is, when $G$ is a direct product of a torus and an adjoint type group).  The problem is that, while the adjoint actions of $G$ on $G_R$ and $\mathfrak{g}_R$ factor through the quotient $G/Z(G)$ by the center, the scheme-theoretic surjection $G \to G/Z(G)$ may not yield a surjection on rational points!  There is a cokernel that lies in a flat $H^1$ group.  Even so, I have not been able to construct an example giving a negative answer to either question.
Naturally, I would appreciate any insights even in the case of fields, or with restricted characteristic.
Edit 9 hours later: L. Spice has helpfully pointed out in a comment that when $R$ is a field, then the answer to question 2 is "yes".  I now see that this is theorem 20.9(ii) in Borel's Linear algebraic groups.  I haven't gone through the details yet, but it looks like the proof in the text can be extended to conjugacy that is Zariski local in $R$.
 A: When $R$ is a field, the answer to Question 1 is Yes (at least in char 0) with the same proof as for Question 2.
For simplicity, we write $G$ for $G_R$, $T$ for $T_R$, etc.

Theorem.
Let $G$ be a split  (connected) reductive group over a field $R$ of characteristic 0, and let $T\subset G$ be a split maximal torus. Write ${\mathfrak g}={\rm Lie\,} G$ and ${\mathfrak t}={\rm Lie\,} T$.
  Let $a\in{\rm Aut\,} {\mathfrak t}$ be an automorphism of ${\mathfrak g}$, and set ${\mathfrak t}'=a({\mathfrak t})\subset {\mathfrak g}$.
  Then there exists $g\in G(R)$ such that ${\mathfrak t}'={\rm Ad}(g)({\mathfrak t})$.

Proof.
First let us canonically decompose ${\mathfrak g}$ and ${\mathfrak t}$.
Write ${\mathfrak z}$ for the center of ${\mathfrak g}$, 
and write ${\mathfrak g}^{\rm ss}=[{\mathfrak g},{\mathfrak g}]$, which is a semisimple Lie algebra.
Then ${\mathfrak g}={\mathfrak z}\dotplus {\mathfrak g}^{\rm ss}$ (direct sum) and hence,
$$ {\rm Aut\,} {\mathfrak g}={\rm Aut\,} {\mathfrak z}\times {\rm Aut\,}{\mathfrak g}^{\rm ss}.$$
Set ${\mathfrak t}^{\rm ss}={\mathfrak t}\cap{\mathfrak g}^{\rm ss}$; then ${\mathfrak t}={\mathfrak z}\dotplus{\mathfrak t}^{\rm ss}$.
Write 
$$a=(a^{\mathfrak z}, a^{\rm ss})\in {\rm Aut\,} {\mathfrak z}\times {\rm Aut\,}{\mathfrak g}^{\rm ss}={\rm Aut\,} {\mathfrak g}.$$
Then it is clear that 
$$a({\mathfrak t})=a^{\mathfrak z}({\mathfrak z})\dotplus a^{\rm ss}({\mathfrak t}^{\rm ss})={\mathfrak z}\dotplus a^{\rm ss}({\mathfrak t}^{\rm ss})=a^{\rm ss}({\mathfrak t}).$$
Therefore, we may and shall assume that $a\in{\rm Aut\,}{\mathfrak g}^{\rm ss}$.
Now let us canonically decompose $G$.
Write $Z^0=Z(G)^0$ (the identity component of the center of $G$).
Then $Z^0$ is a torus with Lie algebra ${\rm Lie\,} Z^0={\mathfrak z}$.
Write $G^{\rm ss}=[G,G]$ (the commutator subgroup of $G$),
which is a connected semisimple group with Lie algebra ${\rm Lie\,} G^{\rm ss}={\mathfrak g}^{\rm ss}$.
Let $G^{\rm sc}$ denote the universal cover of $G^{\rm ss}$.
Then $G^{\rm sc}$ is a simply connected semisimple group with Lie algebra ${\rm Lie\,} G^{\rm sc}={\mathfrak g}^{\rm ss}$.
We have a canonical homomorphism
$$\rho\colon G^{\rm sc}\to G^{\rm ss}\hookrightarrow G$$
and a decomposition (in general not direct)
$$G=Z^0\cdot\rho(G^{\rm sc}).$$
We can canonically decompose $T$. 
Set $T^{\rm ss}=T\cap G^{\rm ss}$, and let $T^{\rm sc}$ denote the preimage of $T^{\rm ss}$ in $G^{\rm sc}$.
Then $T^{\rm ss}$ is a maximal torus in $G^{\rm ss}$, and $T^{\rm sc}$ is a maximal torus in $G^{\rm sc}$.
We have 
$$ T=Z^0\cdot T^{\rm ss}=Z^0\cdot \rho(T^{\rm sc}).$$
We may and shall identify ${\rm Aut\,}{\mathfrak g}^{\rm ss}$ with ${\rm Aut\,} G^{\rm sc}$; then $a\in {\rm Aut\,} G^{\rm sc}$.
Set $T'=Z^0\cdot \rho(a(T^{\rm sc}))\subset G$.
Then $T'$ is a maximal torus in $G$ with Lie algebra 
$${\rm Lie\,} T'={\mathfrak z}\dotplus{\rm Lie\,} a(T^{\rm sc})={\mathfrak z}\dotplus a({\rm Lie\,} T^{\rm sc})={\mathfrak z}\dotplus a({\mathfrak t}^{\rm ss})=a({\mathfrak t})={\mathfrak t}'.$$
Since $T$ is a split torus, the tori $Z^0$, $T^{\rm ss}$, $T^{\rm sc}$, and $a(T^{\rm sc})$ are all split.
Thus $T'$ is a split maximal torus in $G$.
By Theorem 20.9(ii) in Borel's book Linear Algebraic Groups, 
there exists $g\in G(R)$ such that $T'=g\cdot T\cdot g^{-1}$.
Then ${\mathfrak t}'={\rm Ad}(g)({\mathfrak t})$, as required. 
