**Edite** According to the essential comment of Ian Agol I revise the question as follows


For a smooth manifold $M$, is there a non identity involution $\theta$ on the lie algebra $\chi^{\infty}(M)$ such that $X$ is topological equivalent to $\theta (X)$, for all smooth vec. field $X$ on $M$?


This question is motivated by the fact that, on $\mathbb{R}^{n}$ the linear vector field $\dot X=AX$ is topological equivalent to the linear vector field $\dot X=-A^{tr}X$. 


By topological equivalent, we mean existence of an orbit preserving homeomorphism.



A question on involutions on the Lie algebra of vector fields
The Lie algebra of vector fields of a smooth manifold determines the manifold with its smooth structure. Below I indicate a series of papers where this is proved in various settings. In particular, this implies that any Lie algebra involution of the Lie algebra of vector fields has to be induced by an involution of the manifold.


* MR0064764 (16,331a) 
Shanks, M. E.; Pursell, Lyle E.
The Lie algebra of a smooth manifold. 
Proc. Amer. Math. Soc. 5, (1954). 468–472.
* MR0375400 (51 #11594)  

Amemiya, Ichiro; Masuda, Kazuo; Shiga, Kōji
Lie algebras of differential operators. 
Osaka J. Math. 12 (1975), 139–172.
* MR0516602 (80g:57036)  

Grabowski, J.
Isomorphisms and ideals of the Lie algebras of vector fields. 
Invent. Math. 50 (1978/79), no. 1, 13–33.
* MR2785498 (2012b:58028) 
Grabowski, Janusz(PL-PAN); Kotov, Alexei(LUX-CUL); Poncin, Norbert(LUX-CUL)
Geometric structures encoded in the Lie structure of an Atiyah algebroid. (English summary) 
Transform. Groups 16 (2011), no. 1, 137–160


