I will start with the simpler case where the vector field $\mathbf{X}$ is a $C^{\infty}$ vector field on $\mathbb{R}^{2}$. For simplicity, let us assume that $\mathbf{X}$ is non-singular on all of $\mathbb{R}^{2}$ and that it satisfies $\mathbf{X}\cdot\mathbf{X} = 1$, i.e., the vector field $\mathbf{X}$ has been normalized so that the integral curves of $\mathbf{X}$ are parametrized by arc-length. (Here, `$\cdot$' represents the ordinary Euclidean dot product. I will make every effort to keep the response in terms of ordinary Euclidean geometry, although some tensor analysis could possibly simplify the exposition.)
Write $\mathbf{X}$ in standard Cartesian coordinates as
$$
\mathbf{X}(x, y) = f(x, y)\frac{\partial}{\partial x} + g(x, y) \frac{\partial}{\partial y} = \begin{pmatrix}f(x, y)\\ g(x, y)\end{pmatrix}.
$$
(I will employ the convention of vector coordinates relative to a basis being column vectors.)
Now consider a point $P = P(x_{0}, y_{0})$ in the plane and denote the integral curve of $\mathbf{X}$ passing through $P$ at time zero by $\alpha(s) = (x(s), y(s))$.
By definition we then have that $\alpha(0) = P$ and
$$\frac{d\alpha}{ds} = \mathbf{X}\circ\alpha(s) = \begin{pmatrix}f\left(\alpha\left(s\right)\right)\\ g\left(\alpha\left(s\right)\right)\end{pmatrix}.$$
Note that our assumptions on the vector field $\mathbf{X}$ imply that $\frac{d\alpha}{ds}\cdot \frac{d\alpha}{ds} = 1$. It follows that the unit tangent vector field along our curve $\alpha$ is
$$
\mathbf{T}(s) = \frac{d\alpha}{ds} = \mathbf{X}\circ\alpha(s) = \begin{pmatrix}f\left(\alpha\left(s\right)\right)\\ g\left(\alpha\left(s\right)\right)
\end{pmatrix},
$$
while the (oriented) unit normal vector field $\mathbf{N}$ along $\alpha$ is obtained by a positive counterclockwise rotation of $\pi/2$ radians and is given in coordinates by
$$
\mathbf{N}(s) = \begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix} \begin{pmatrix}f\left(\alpha\left(s\right)\right)\\ g\left(\alpha\left(s\right)\right)\end{pmatrix} = \begin{pmatrix}-g\left(\alpha\left(s\right)\right)\\ f\left(\alpha\left(s\right)\right)\end{pmatrix}
$$
Standard properties of differentiation and the dot product give that the vector field $\frac{d\mathbf{T}}{ds}$ along $\alpha$ is perpendicular to $\mathbf{T}$ and a scalar multiple of $\mathbf{N}$. The (oriented) curvature function $\kappa$ along $\alpha$ is then found by differentiating the vector field $\mathbf{T}$ and the relation
$$
\frac{d\mathbf{T}}{ds} = \frac{d^{2}\alpha}{ds^2} = \kappa(s)\mathbf{N}(s).
$$
Since $\mathbf{T}(s) = \frac{d\alpha}{ds} = \mathbf{X}\circ \alpha(s)$ we have the following: (To avoid a proliferation of parentheses, all functions and vector fields defined on $\mathbb{R}^{2}$ are assumed to be evaluated along the integral curve $\alpha(s)$.)
\begin{align*}
\frac{d\mathbf{T}}{ds} &= \frac{d}{ds}\left(\mathbf{X}\circ\alpha(s)\right)
= \begin{pmatrix}
\nabla f \cdot \mathbf{X} \\
\nabla g \cdot \mathbf{X}
\end{pmatrix}
=
\nabla \mathbf{X}\bullet \mathbf{X}\\
\end{align*}
On the last expression in the string of equalities above, $\displaystyle \nabla \mathbf{X} = \begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\\
\frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\\
\end{pmatrix}$ is the Jacobian of the vector field $\mathbf{X}$ and $\nabla \mathbf{X}\bullet \mathbf{X}$ understood to be matrix multiplication. (Again, for emphasis, all functions/vector fields on $\mathbb{R}^{2}$ are understood to be evaluated along the integral curve $\alpha$.)
At this point, one can solve the equation
$$
\frac{d\mathbf{T}}{ds} = \begin{pmatrix}
\nabla f \cdot \mathbf{X} \\
\nabla g \cdot \mathbf{X}
\end{pmatrix}
= \kappa(s)
\begin{pmatrix}
-g\\
f\\
\end{pmatrix}
$$
for the (oriented) curvature function $\kappa$, or use a well-known identity for unit-speed curves in the plane such as
$$
\kappa(s) = \det \left(\frac{d\alpha}{ds}, \frac{d^2\alpha}{ds^2}\right) = \det\begin{pmatrix} f & \nabla f\cdot \mathbf{X}\\ g & \nabla g \cdot \mathbf{X}\end{pmatrix}.
$$
Either way, one finds
$$
\kappa(s) = \left(f \nabla g - g \nabla f\right)\cdot \mathbf{X},
$$
where the right hand side is evaluated along the integral curve $\alpha(s)$.
We now turn our attention to the case where $\mathbf{X}$ is a vector field on the three-dimensional Euclidean space $\mathbb{R}^{3}$. As before, we will assume that the vector field $\mathbf{X}$ satisfies $\mathbf{X}\cdot\mathbf{X} = 1$.
We will express $\mathbf{X}$ in the standard coordinate frame as
$$
\mathbf{X}\left(x, y, z\right) = f(x, y, z) \frac{\partial }{\partial x} + g(x, y, z) \frac{\partial }{\partial y} + h(x, y, z) \frac{\partial }{\partial z} =
\begin{pmatrix}
f(x, y, z)\\
g(x, y, z)\\
h(x, y, z)\\
\end{pmatrix}.
$$
Let $\alpha : \mathbb{R} \to \mathbb{R}^{3}$ be the integral curve of $\mathbf{X}$ that passes through $\displaystyle P = P\left(x_{0}, y_{0}, z_{0}\right)$ at time $s = 0$. As before it follows that
$$
\frac{d\alpha}{ds} = \mathbf{T}(s) = \mathbf{X}\circ \alpha(s)
$$
is the unit tangent vector field along $\alpha$.
Standard considerations give that $\mathbf{T}$ is perpendicular to $\mathbf{T}^\prime = \frac{d\mathbf{T}}{ds}$. Assuming that $\mathbf{T}^{\prime}(s)$ is non-zero (which is at least the case on an appropriate open interval about 0 when $\mathbf{T}'(0)$ is not the zero vector and the integral curve is not a line/geodesic), then the unit normal vector field $\mathbf{N}$ along $\alpha$ is defined by
$$
\mathbf{N} = \frac{\mathbf{T}^{\prime}}{\lvert\lvert \mathbf{T}^{\prime}\rvert\rvert},
$$
with the curvature $\displaystyle \kappa(s) = \lvert\lvert \mathbf{T}^{\prime}\rvert\rvert$.
A similar calculation to the case when the vector field $\mathbf{X}$ is a vector field on the plane gives
\begin{equation}
\frac{d\mathbf{T}}{ds} = \frac{d}{ds}\left(\mathbf{X}\circ\alpha(s)\right)
= \begin{pmatrix}
\nabla f \cdot \mathbf{X} \\
\nabla g \cdot \mathbf{X}\\
\nabla h \cdot \mathbf{X}
\end{pmatrix}
=
\nabla \mathbf{X}\bullet \mathbf{X},\\
\end{equation}
where $\displaystyle \nabla \mathbf{X} = \begin{pmatrix} \nabla f & \nabla g &\nabla h \end{pmatrix}^{t}$ is the Jacobian of $\mathbf{X}$ and $\nabla \mathbf{X}\bullet \mathbf{X}$ is matrix multiplication. (Again, everything defined on all of $\mathbb{R}^{3}$ is assumed to be evaluated along the integral curve $\alpha$. )
The curvature function $\kappa$ of the integral curve $\alpha$ is then given by
$$
\kappa(s) = \sqrt{\left(\nabla \mathbf{X}\bullet \mathbf{X}\right) \cdot \left(\nabla \mathbf{X}\bullet \mathbf{X}\right)}.
$$
The unit binormal vector field $\mathbf{B}$ along $\alpha$ is defined by $\mathbf{B} = \mathbf{T} \times \mathbf{N}$.
Identifying
$$ \mathbf{N} = \frac{\mathbf{T}^\prime}{\lvert\lvert \mathbf{T}^{\prime}\rvert\rvert} = \frac{1}{\kappa} \begin{pmatrix}
\nabla f \cdot \mathbf{X} \\
\nabla g \cdot \mathbf{X}\\
\nabla h \cdot \mathbf{X}
\end{pmatrix},$$
a cross-product calculation gives
\begin{align*}
\mathbf{B} &= \frac{1}{\kappa}
\begin{pmatrix}
g \nabla h \cdot \mathbf{X} - h\nabla g \cdot\mathbf{X}\\
h\nabla f\cdot \mathbf{X} - f \nabla h \cdot \mathbf{X}\\
f \nabla g\cdot\mathbf{X} - g \nabla f \cdot \mathbf{X}
\end{pmatrix}\\
&=
\frac{1}{\kappa}
\begin{pmatrix}
\left(g \nabla h - h\nabla g\right) \cdot\mathbf{X}\\
\left(h\nabla f- f \nabla h\right) \cdot \mathbf{X}\\
\left(f \nabla g - g \nabla f\right) \cdot \mathbf{X}
\end{pmatrix}\\
&= \frac{1}{\kappa}
\begin{pmatrix}
\left(g \nabla h - h\nabla g\right)^{t} \\
\left(h\nabla f- f \nabla h\right)^{t} \\
\left(f \nabla g - g \nabla f\right)^{t}
\end{pmatrix}
\bullet \mathbf{X}
\end{align*}
At this point, it is not clear to me the most efficient way to calculate the torsion function $\tau$ of the integral curve $\alpha$, although one option is as follows. For a unit-speed curve $\alpha$ with non-zero curvature in three-dimensional Euclidean space, we have that the torsion $\tau$ is
\begin{align*}
\tau &= \frac{1}{\kappa^2} \left(\mathbf{T} \times \mathbf{T}^{\prime}\right)\cdot \mathbf{T}^{\prime\prime}\\
&\\
&=\frac{1}{\kappa}\mathbf{B}\cdot\mathbf{T}^{\prime\prime}.
\end{align*}
Additionally, we have everything except $\mathbf{T}^{\prime\prime}$ expressed in terms of the components s the vector field $\mathbf{X}$. Differentiating (1) with respect to $s$ yields
$$
\frac{d}{ds}\left(\frac{d\mathbf{T}}{ds}\right) = \mathbf{T}^{\prime\prime} = \begin{pmatrix}
\left(f\nabla f_{x} + f_{x} \nabla f + g \nabla f_{y} + f_{y} \nabla g + h \nabla f_{z} + f_{z}\nabla h \right)\cdot \mathbf{X}\\
\left(f\nabla g_{x} + g_{x} \nabla f + g \nabla g_{y} + g_{y} \nabla g + h \nabla g_{z} + g_{z}\nabla h \right)\cdot \mathbf{X}\\
\left(f\nabla h_{x} + h_{x} \nabla f + g \nabla h_{y} + h_{y} \nabla g + h \nabla h_{z} + h_{z}\nabla h \right)\cdot \mathbf{X}
\end{pmatrix}
$$
As before, all expressions above are evaluated along the integral curve $\alpha = \alpha(s)$.
(It would be nice if the expression for $\mathbf{T}^{\prime\prime}$ could be succinctly summarized using some vector identities, but
I do not see it at the moment.)
One can then use the indicated expressions to calculate the torsion of the integral curve $\alpha$ entirely in terms of the vector field $\mathbf{X}$.
I'll stick to the convention that $x$ is a column vector so that matrices appear on the left, which seems contrary to some parts of this post. The more difficult direction of the two (I think) is to show that if a linear homomorphism $T$ exists, then $A$ and $B$ must satisfy $B = TAT^{-1}$.
So, suppose that $T$ is such that for all $x \in X = \Bbb C^n$ and $t \in \Bbb R$, we have
$$
T\phi_A(x,t) = \phi_B(Tx,t) \tag{1}
$$
where
$$
\phi_A(x,t) = e^{At}x, \qquad \phi_B(x,t) = e^{Bt}x. \tag{2}
$$
Plugging (2) into (1) yields the equation
$$
Te^{At} x = e^{Bt}Tx, \tag{3}
$$
which holds for all $x \in X$ and $t \in \Bbb R$. Because (3) holds for all $x$, we see that for any given $t \in \Bbb R$, $Te^{At}, e^{Bt}T$ induce the same linear operator which means that they must be the same matrix. That is, we have $Te^{At} = e^{Bt}T$ for all $t$.
Now, note that
$$
\begin{align}
Te^{At} &= T e^{T^{-1}(TAT^{-1}t)T} =
T [T^{-1}e^{TAT^{-1}t}T] = e^{(TAT^{-1})t} T.
\end{align}
$$
So, we can now state that $e^{(TAT^{-1})t} T = e^{Bt}T$ holds for all $t$. By the invertibility of $T$, this implies that $e^{(TAT^{-1})t} = e^{Bt}$ holds for all $t$. It can now seen in a few ways that $TAT^{-1} = B$; I think that the most natural is to note that
$$
\left. \frac{d}{dt} e^{Mt} \right|_{t=0} = M e^{M\cdot 0} = M.
$$
So since the functions $e^{(TAT^{-1})t}, e^{Bt}$ are equal for all $t$, they're derivatives at $t = 0$ must coincide so that $TAT^{-1} = B$, as was desired.
Best Answer
As Didier commented above, in the diffeomorphism case there is a simple relationship; I would like to highlight the "space of vector fields is the Lie algebra of the diffeomorphism group" perspective (see Is there an infinite dimensional Lie group associated to the Lie algebra of all vector fields on a manifold? , Is there a way to associate a Lie algebra to the group of diffeomorphisms?):
Let $M$ be a $C^\infty$ manifold, for brevity assume $M$ is compact (otherwise (e.g. in the case of $M=\mathbb{R}^d$ as in the OP) one could focus on compactly supported vector fields and diffeomorphisms, or simply use local flows (or reparametrize local flows to get global flows (see the Hart paper mentioned below), ...). Denote by $\mathcal{V}^r(M),\mathcal{F}^r(M), \mathcal{D}^r(M)$ the spaces of $C^r$ ($r\in\mathbb{Z}_{\geq1}\cup\{\infty\}$) vector fields, flows (i.e. $\mathbb{R}$-actions), and diffeomorphisms, respectively. $\mathcal{D}^r(M)$ acts on $\mathcal{F}^r(M)$ by conjugation:
$$\Gamma:\mathcal{D}^r(M)\times\mathcal{F}^r(M)\to \mathcal{F}^r(M),\,\, (f,\phi_\bullet)\mapsto [(t,x)\mapsto f\circ\phi_t\circ f^{-1}(x)].$$
$\mathcal{D}^r(M)$ acts on $\mathcal{V}^{r-1}(M)$ by the adjoint operator:
$$\operatorname{Ad}:\mathcal{D}^r(M)\times\mathcal{V}^{r-1}(M)\to \mathcal{V}^{r-1}(M),\,\, (f,X)\mapsto Tf\circ X\circ f^{-1}.$$
Evaluating the time derivative at $0$ gives a map $\left.\frac{\partial}{\partial t}\right|_{t=0}:\mathcal{F}^r(M)\to \mathcal{V}^{r-1}(M)$:
$$[\phi: \mathbb{R}\times M\to M]\mapsto \left[M\to TM,\,\,x\mapsto \left.\dfrac{\partial\phi}{\partial t}\right|_{t=0}(t,x)\right],$$
and conversely the Existence and Uniqueness Theorem in ODE's says that there is a unique map $\phi^\bullet:\mathcal{V}^r(M)\to \mathcal{F}^r(M)$ such that
$$\left.\dfrac{\partial \phi^X}{\partial t}\right|_{t=0}(t,x)=X(x).$$
(Note that the notation $\phi^X(t,x)=\exp(tX)(x)$ is also in use; compare Why is the domain of the exponential function the Lie algebra and not the Lie group?)
Thus we have:
Claim: If $X\in\mathcal{V}^r(M)$ and $f\in\mathcal{D}^{r+1}(M)$, then $\left.\dfrac{\partial}{\partial t}\right|_{t=0}\Gamma_f(\phi^X)=\operatorname{Ad}_f(X)\in\mathcal{V}^r(M)$, or equivalently, $\phi^{\operatorname{Ad}_f(X)}=\Gamma_f(\phi^X)\in\mathcal{F}^r(M)$.
Proof:
\begin{align*} \left.\dfrac{\partial}{\partial t}\right|_{t=0}\Gamma_f(\phi^X)(x) &=\left.\dfrac{\partial}{\partial t}\right|_{t=0}f(\phi^X(t,f^{-1}(x)))\\ &= T_{\phi^X(0,f^{-1}(x))}f \,\,\,\left.\dfrac{\partial\phi^X}{\partial t}\right|_{t=0}(t,f^{-1}(x)))\\ &= T_{f^{-1}(x)}f\,\, X(f^{-1}(x)) =Tf\circ X\circ f^{-1}(x) =\operatorname{Ad}_f(X)(x). \end{align*}
Thus we have a diagram
In the case when the conjugation is by a homeomorphism for obvious reasons the above does not work. Indeed, in this case the conjugacy may break all infinitesimal structure, and it is reasonable to look at other properties that are invariant; e.g. under a topological conjugacy the time durations of corresponding periodic orbits will be the same. Similarly limit sets will be preserved. In fact, that there is no such simple relation in the homeomorphism case can be considered as one of the reasons why dynamics is interested in topological conjugacy (or other versions of it, like topological conjugacy on nonwandering sets, or orbit equivalence (where the equivalence does not necessarily preserve the time parameter)); differentiable conjugacy is too rigid. (See Smale's "Differentiable Dynamical Systems" (pp.748-749) of Hirsch, Smale & Devaney's Differential Equations, Dynamical Systems, and an Introduction to Chaos (3e) (p.66) for more on this point.) It is also interesting to consider which properties of a differentiable conjugacy would allow one to upgrade a topological conjugacy (this is the study of "moduli of conjugacy" or "moduli of stability", e.g. see Palis' "A Differentiable Invariant of Topological Conjugacies and Moduli of Stability").
Let me also mention a result by Hart that fits well into the framework above. Note that in general we have the following regularity relations (for $p,q,r,s\in\mathbb{Z}_{\geq1}\cup \{\infty\}$):
$$\Gamma:\mathcal{D}^r(M)\times\mathcal{F}^p(M)\to \mathcal{F}^{\min\{r,p\}}(M),$$
$$\operatorname{Ad}:\mathcal{D}^s(M)\times\mathcal{V}^q(M)\to \mathcal{V}^{\min\{s-1,q\}}(M).$$
As mentioned above one can differentiate a flow in the time variable to get a vector field, but the regularity may drop. Indeed if $X\in \mathcal{V}^q(M)$ and $f\in\mathcal{D}^q(M)\setminus \mathcal{D}^{q+1}(M)$, then $\operatorname{Ad}_f(X)\in \mathcal{V}^{q-1}(M)\setminus \mathcal{V}^q(M)$, but the flow of this vector field (based on the above calculation) is $\phi^{\operatorname{Ad}_f(X)}=\Gamma_f(\phi^X)\in\mathcal{F}^q(M)$; so that we have a $C^q$ flow whose generator is not $C^q$.
Theorem (Hart): Let $M$ be a $C^\infty$ manifold (not necessarily compact). For any $\phi\in\mathcal{F}^r(M)$ and for any $\epsilon\in\mathbb{R}_{>0}$, there is a $g\in\mathcal{D}^r(M)$ such that $d_{C^r}(g,\operatorname{id}_M)<\epsilon$ and $\left.\frac{\partial}{\partial t}\right|_{t=0} \Gamma_g(\phi)\in\mathcal{V}^r(M)$.
(See Hart's paper "On the Smoothness of Generators" for this and an analogous result for foliations; $d_{C^r}$ is the distance of uniform $C^r$ convergence on compact subsets; see e.g. Hirsch's Differential Topology vs Rudin Functional analysis definition of weak and strong topology.)