Similar Matrices and Conjugate Flows This question is an attempt to resuscitate and generalize this question, which lived one hour and was deleted by its author for reasons unknown.
The notion of topological conjugacy of flows is central to the theories of dynamical systems and ordinary differential equations, for it formalizes the concept that the orbit structure of two flows may be equivalent; that is, that the flows exhibit essentially the same behavior.  Specifically, we say that two flows $\phi_A$ and $\phi_B$ on a topological space $X$ are conjugate when there exists a homeomorphism
$h:X \to X \tag 1$
such that for all $x \in X$ and $t \in \Bbb R$
$h(\phi_A(x, t)) = \phi_B(h(x), t). \tag 2$
A simple and particularly useful class of flows arises from linear time-invariant ordinary differential equations
$\dot x = Ax, \tag 3$
where we assume
$x \in \Bbb C^n \tag 4$
and
$A \in M(n, \Bbb C), \tag 5$
the set of square complex matrices of size $n$; the solution to (3) which takes the value $x(t_0)$ at $t_0$ is well-known to be
$x(t) = e^{A(t - t_0)}x(t_0); \tag 6$
thus the flow of (3) is given by
$\phi_A(x, t) = e^{At} x. \tag 7$
We recall that two matrices
$A, B \in M(n, \Bbb C) \tag 8$
are said to be similar if there exists a non-singular matrix
$T \in M(n, \Bbb C) \tag 9$
such that
$B = TAT^{-1}. \tag{10}$
The Question is then:
Show that there is a linear homeomorphism
$T: \Bbb C^n \to \Bbb C^n \tag{11}$
conjugating $\phi_A(x, t)$ and $\phi_B(x, t)$ if and only if $A$ and $B$ are $T$-similar, that is, 
$B = TAT^{-1}. \tag{12}$
If $A$ and $B$ are real matrices, must $T$ also be real?
Note Added in Edit, Saturday 8 February 2020, 11:25 AM PST:  In light of Conifold's comment, we see that the existence of a suitable real $T$ gives rise to a purely imaginary matrix $iT$ which also satisfies the requisite conditions.  Therefore I modify my closing question above and ask 
If $A$ and $B$ are real matrices, must $T$ be either purely real or purely imaginary?
End of Note.
 A: 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.
