Complex matrices are similar over $R$ iff they and thier conjugates are similar over $C$. 
Prove that there is a nonsingular $T \in M_n(R)$ such that $A = TBT^{−1}$ if and only if there is a nonsingular $S \in M_n$ such that both $A = SBS^{−1}$ and $\bar{A} = S\bar{B}S^{−1}$. (Here $M_n$ is the set of $n \times n$ complex matrices).

If direction: We can take $S = T$ i.e. imaginary part being zero. Then, $A = SBS^{-1}$ obviously. Moreover, $\bar{A} = \overline{SBS^{-1}} = \bar{S}\bar{B}\bar{S^{-1}} = S\bar{B}S^{-1}$.
The only if direction: Let $S = H+iC$ and $\bar{S} = H-iC$ where $H,C \in M_n(R)$. Then,
$A = SBS^{-1} = (H + iC)B(H+iC)^{-1}, \bar{A} = \bar{S}\bar{B}\bar{S^{-1}} = (H-iC)\bar{B}(H-iC)^{-1} = S\bar{B}S^{−1} =(H + iC)\bar{B}(H+iC)^{-1}$.
I guess I should show that $S = \bar{S}$ i.e. $H+iC = H - iC$. Because if complex conjugate of $S$ is equal to $S$ then $S \in M_n(R)$ so, we will complete the proof.
 A: The only direction you need help with here is the "only if" direction. We are given that
$$
A = SBS^{-1}, \quad  \bar A = S\bar BS^{-1}.
$$
Rearranging these equations and taking the conjugate of both sides on the right brings us to the equations
$$
AS = SB, \quad A \bar S = \bar S B.
$$
We can therefore conclude that $H$ and $C$ (the real and imaginary parts of $S$) satisfy $AH = HB$ and $AC = CB$. The catch, however, is that we need an invertible real matrix $T$ for which $AT = TB$, and we have no guarantee that either $H$ or $C$ is invertible.
With that in mind, consider the polynomial
$$
p(t) = \det(H + Ct).
$$
We know that $p(i) \neq 0$, which means that $p$ cannot be the zero-polynomial. Thus, $p$ has at most finitely many zeros. Thus, there necessarily exists a $t \in \Bbb R \subset \Bbb C$ for which $p(t) \neq 0$. If we define $T = H + Ct$, then we see that $T$ is real, $T$ is invertible, and $AT = TB$. We can rearrange this last equation to get
$$
A = TBT^{-1},
$$
which was what we wanted.
