In what sense are Minkowski spaces with $(1,3)$ and $(3,1)$ signature isomorphic? There is no isometry between a $(1,3)$-signatured and a $(3,1)$-signatured Minkowski space, but in spite of this, they "look like" the same, for example, they have the same light cone. Is there any treatment (category) which defines a special kind of isomorphism so that $(1,3)$ and $(3,1)$-signatured Mikowski spaces are isomorhic (or the same)?
Edit: perhaps we should regard the Zeeman-topologies?
 A: Suppose $(V_1, q_1), (V_2, q_2)$ are $4$-dimensional vector spaces equipped with quadratic forms $q_1, q_2$ of signatures $(3,1)$ and $(1,3)$ respectively. Let $G_1, G_2$ denote the automorphism groups of the forms $q_1, q_2$, i.e. of linear transformations preserving the respective forms. Then the groups $G_1, G_2$ are isomorphic ($O(3,1)\cong O(1,3)$). More precisely, there exists an invertible linear map $L: V_1\to V_2$ such that:


*

*$L_*(q_1)=-q_2$, i.e. for every vector $v\in V_1$ we have
$$
q_2(L(v))= -q_1(v). 
$$

*$L G_1 L^{-1}= G_2$. 
Hence, the linear map $L$ preserves everything we care about when working with Lorentzian geometry. 
One more thing: The same works for vector spaces equipped with quadratic forms of signatures $(p,q)$ and $(q,p)$, for instance, you can take the form $q_1$ to be positive-definite and $q_2$ negative-definite. 
As for complexifying: This would be cheating since after that even forms of signatures $(3,1)$ and $(2,2)$ become isomorphic.  
A: Consider the Minkowski space as $\mathbb{R}^4$ and equip it with the bilinear form $B(x,y)=x^tAy$ where $$A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$
Definition: A vector space equipped with a quadratic form is called a quadratic space. In geometry, we like inner product spaces and they come equipped with a bilinear form that satisfies some extra properties like non-negativity, symmetry and non-degeneracy. Once we have a bilinear form $B$ on a vector space, we can induce a quadratic form $q: V\to \mathbb{F}$ on our vector space simply by considering $q(x)=B(x,x)$. 
Now two bilinear forms $B_1$ and $B_2$ over an $n$-dimensional vector space over $\mathbb{F}$ are called isomorphic when there exists an invertible transformation $T\in \mathrm{GL}_n(\mathbb{F})$ such that $B_2(v,w)=B_1(Tv,Tw)$. Likewise, they're called anti-isomorphic if we instead have $B_2(v,w)=-B_1(Tv,Tw)$. You can formulate these definitions for quadratic forms as well.
In our case, it's easy to see that the different signatures $(3,1)$ and $(1,3)$ have quadratic forms that are anti-isomorphic by taking $T=I$. However, thanks to Moishe Cohen's answer, their automorphism groups turns out to be isomorphic by conjugation where conjugation is done using an anti-isomorphism. You can now see that the two version of the Poincare group defined using the two different signatures are isomorphic.
