Conjugation in finite simple classical groups Let $G_n(q)=\mathrm{(P)SL}_n(q), \mathrm{(P)SU}_n(q),\mathrm{(P)Sp}_{2n}(q)$, $\Omega_{2n+1}(q), (P)\Omega^\pm_{2n}(q)$ be a simple classical group. Consider the natural embedding $G_{n-1}(q) \subset G_n(q)$ if there is such. I would appreciate very much if you have an argument or a counterexample for the following:
Claim: If $x,y\in G_{n-1}(q)$ such that $x^T=y$ for some $T\in \mathrm{Aut}(G_n(q))$, then there exists $t\in \mathrm{Aut}(G_{n-1}(q))$ such that $x^t=y$. 
P/S: I have modified the question slightly based on Nick's comment.
 A: [Again misread the question... sorry. Leaving the answer for context.]
A bit sketchy, but I guess it can be filled.
Consider $\mathrm{SO}(8)$, with respect with the quadratic form $J_8$ given by the antidiagonal matrix. A Cartan subalgebra consists of the matrices $\mathrm{diag}(x,y,z,t,-t,-z,-y,-x)$. Choose these coordinates, for which the Killing form on the Cartan subalgebra is proportional to the usual scalar product. Then (up to a single scalar), the simple coroots (for the positivity corresponding to upper triangular matrices) are $\alpha_1=(0,1,-1,0)$, $\alpha_2=(1,-1,0,0)$, $\alpha_3=(0,0,1,-1)$, and $\alpha_4=(0,0,1,1)$. The last three are pairwise orthogonal, and triality permutes them arbitrarily. In particular, some automorphism maps $\alpha_2+\alpha_3=(1,-1,1,-1)$ to $\alpha_3+\alpha_4=(0,0,2,0)$. That is, at a group level, it maps $\mathrm{diag}(t,t^{-1},t,t^{-1},t,t^{-1},t,t^{-1})$ to $\mathrm{diag}(1,1,t^2,1,1,t^{-2},1,1)$. For $t\neq \pm 1$, these are not linearly conjugate (even in $\mathrm{PSO}_8$). 
Based on this, I believe that for $t\neq\pm 1$ in any finite field $K$ of odd cardinal, except maybe a few exceptions, that $\mathrm{diag}(t,t^{-1},t,t^{-1},t,t^{-1},t,t^{-1})$ (which modulo scalar is $\mathrm{diag}(t^2,1,t^2,1,t^2,1,t^2,1)$) and $\mathrm{diag}(1,1,t^2,1,1,t^{-2},1,1)$ are conjugate (by an automorphism) in $\mathrm{PSO}_8(K)$, but are not conjugate (by any automorphism) in a larger $\mathrm{SO}_n$ (with respect to a form $J_8\oplus J_{n-8}$), since for $n\ge 9$ the automorphisms should come from $\mathrm{O}_n$ and field automorphisms.
Note: An easier non-Lie analogue (of this phenomenon, where the question is reversed): 3-cycles and double 3-cycles are automophism-conjugate in $\mathrm{Alt}_6$, but not in $\mathrm{Alt}_n$ for $n\ge 7$.
