= Solution
Write the complex similarity <matrix> as
$$
S=X+iY,
$$
with $X,Y$ real. From $C=SDS^{-1}$ we obtain $CS=SD$, and equating real and imaginary parts gives
$$
CX=XD,\qquad CY=YD.
$$
Therefore
$$
C(X+tY)=(X+tY)D
$$
for every real $t$.
Now $p(t)=\det(X+tY)$ is a real <polynomial> and
$$
p(i)=\det(X+iY)=\det S\ne0.
$$
Thus some real $t$ has $p(t)\ne0$. Taking $T=X+tY$, we have a real invertible <matrix> satisfying $CT=TD$, or
$$
\boxed{C=TDT^{-1}}.
$$
This proves <real similarity from complex similarity>.
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