 Solution

The matrix trace is $\mathrm{Tr}A={\sum }_i{A}_{ii}$. Expanding the matrix products gives $\mathrm{Tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}={\sum }_{j,i}{B}_{ji}{A}_{ij}=\mathrm{Tr}(BA)$. Thus the cyclic property of the trace gives zero trace to every commutator. If $AB-BA=I$, taking the trace would give $0=n$, impossible for positive finite $n$ over the real or complex field.

For real $A,B$, let $p(\lambda )=\det (A+\lambda B)$, a real polynomial of degree at most $n$. Since $p(i)=\det C\ne 0$, it is not the zero polynomial. It has at most finitely many real roots of a polynomial, so $A+\lambda B$ is an invertible matrix for some real $\lambda$, in fact for all but finitely many.

Finally write the invertible intertwiner as $R=A+iB$. The relation $PR=RQ$ and the reality of $P,Q$ give $PA=AQ$ and $PB=BQ$. Choose a real $\lambda$ for which $S=A+\lambda B$ is invertible. Then $PS=SQ$, so $P=SQ{S}^{-1}$. This proves that complex similarity of real matrices implies real similarity.