Yes. Quoting Halmos's Linear algebra problem book (Solution 160).
“If $A$ and $B$ are real, $U$ is unitary, and $U^*AU = B$, then there exists a real orthogonal $V$ such that $V^*AV = B$.
A surprisingly important tool in the proof is the observation that the unitary equivalence of $A$ and $B$ via $U$ implies the same result for $A^*$ and $B^*$. Indeed, the adjoint of the assumed equation is $U^*A^*U = B^*$.
Write $U$ in terms of its real and imaginary parts $U = E + i F$. It follows from $AU = UB$ that $AE = EB$ and $AF = FB$, and hence that $A(E+\lambda F) = (E+\lambda F)B$ for every scalar $\lambda$. If $\lambda$ is real and different from a finite number of troublesome scalars (the ones for which $\det(E+\lambda F) = 0$), the real matrix $S = E + \lambda F$ is invertible, and, of course, has the property that $AS=SB$.
Proceed in the same way from $U^*A^*U = B^*$: deduce that $A^*(E+\lambda F) = (E+\lambda F)B^*$ for all $\lambda$, and, in particular, for the ones for which $E+\lambda F$ is invertible, and infer that $A^*S = SB^*$ (and hence that $S^*A = BS^*$).
Let $S =VP$ be the polar decomposition of $S$ (that theorem works just as well in the real case as in the complex case, so that $V$ and $P$ are real.) Since $$BP^2 = BS^*S = S^*AS = S^*SB = P^2B,$$ so that $P^2$ commutes with $B$, it follows that $P$ commutes with $B$. Since $$AVP = AS = SB = VPB = VBP$$ and $P$ is invertible, it follows that $AV=VB$, and the proof is complete.”
Needless to say, that isn't the shortest path to prove the reduction of antisymmetric matrices...
Best Answer
The answer to your (imprecise) question lies in the Spectral theorem for normal matrices: normal matrices are precisely those that are unitarily diagonalizable. Hermitian and unitary matrices are special cases: hermitian matrices are normal with real eigenvalues, while unitary matrices are normal with complex eigenvalues of modulus one.
Therefore to answer your question, you should look for some matrix with complex non-real and non-unitary eigenvalues.