Consider an antilinear involution, that is an antilinear map on a complex vector space, whose matrix $M$ obeys $MM^\*=1$ where the star denotes complex conjugation. Can we find a change of basis whose matrix $\Lambda$ would be such that $\Lambda^\* M \Lambda^{-1} = 1$? 


By taking the real components of $M$ and $\Lambda$, this can be reduced to a special case of the following problem: given four real commuting square matrices of the same size $A,B,C,D$ such that $AD=BC$, do real vectors $X$ and $Y$ such that $AX=BY$ and $CX=DY$ span the whole space? 


The motivation for this question comes from quantum mechanics, where Hermitian conjugation is a antilinear involution on the space of operators. Trivializing this involution means finding a Hermitian basis of operators. 



Can any antilinear involution be trivialized by a change of basis?
If you let $I$ denote multiplication by $\sqrt{-1}$, then the two operators $I$ and $M$ on your vector space (say, $V$) satisfy
$$
I^2 = -1,\qquad M^2 = 1,\qquad\text{and}\qquad IM=-MI.
$$
(The former since $M$ is an involution; the latter follows since $M$ is anti-linear.) The operators $1,I,M, IM$ span an algebra isomorphic to $M\_2(\mathbb{R})$, the ring of $2$-by-$2$ matrices with real entries, and thus, $V$ is a left module over $M\_2(\mathbb{R})$.


It is well-known that any finite dimensional left module over $M\_2(\mathbb{R})$ is a direct sum of a finite number of copies of the standard $2$-dimensional module $\mathbb{R}^2$ (with its natural action). Thus, your $V$ has a basis that puts it in standard form as a direct sum of copies of $\mathbb{R}^2$. 


I believe this basis provides the matrix $\Lambda$ that you seek.


