The complex structures A² =-I on form exactly two components

Idea. A matrix $A$ with $A^2=-I$ makes $\mathbb{R}^n$ into a complex vector space (multiplication by $i$ is $A$). Complex bases carry a canonical real orientation; comparing it with the standard orientation gives a sign $\varepsilon(A)$, which turns out to be locally constant on $K$.

Step 1: normal form. If $A^2=-I$ then $(\det A)^2=\det(-I)=(-1)^n$, so $n$ is even. Let $A\in K$. Make $V=\mathbb{R}^n$ a complex vector space $V_A$ by $(s+it)x=sx+tAx$ (this is a complex structure because $A^2=-I$). Its complex dimension is $m=n/2$. Choose a $\mathbb{C}$-basis $v_1,\dots,v_m$ of $V_A$. Then $(v_1,Av_1,\dots,v_m,Av_m)$ is a real basis of $\mathbb{R}^n$ (the real span of $v_k,Av_k$ is the complex span of $v_k$). Let $P$ be the matrix with these columns. Then $Av_k=Pe_{2k}$ and $A(Av_k)=-v_k$, i.e. $APe_{2k-1}=Pe_{2k}$ and $APe_{2k}=-Pe_{2k-1}$. Thus
$$AP=PJ_0,\qquad A=PJ_0P^{-1}.$$

Step 2: the sign $\varepsilon(A)$. For $A\in K$ and a list $v=(v_1,\dots,v_m)$ that is a $\mathbb{C}$-basis of $V_A$, put
$$B_A(v)=[\,v_1,Av_1,\dots,v_m,Av_m\,]\in GL_n(\mathbb{R}),\qquad\varepsilon(A)=\operatorname{sign}\det B_A(v).$$
*This does not depend on $v$.* If $v'$ is another $\mathbb{C}$-basis, $v'=G\cdot v$ for a unique $G\in GL_m(\mathbb{C})$, and then $B_A(v')=B_A(v)\,\rho(G)$, where $\rho(G)\in GL_n(\mathbb{R})$ is the "realification" of $G$ (it replaces each entry $g=x+iy$ by the block $\begin{pmatrix}x&-y\\y&x\end{pmatrix}$). The function $G\mapsto\det\rho(G)$ is continuous and nowhere zero on $GL_m(\mathbb{C})$, which is connected, and it equals $1$ at $G=I$. So it is positive everywhere (in fact $\det\rho(G)=|\det G|^2$), and $\operatorname{sign}\det B_A(v')=\operatorname{sign}\det B_A(v)$.

Step 3: $\varepsilon$ is locally constant. Fix $A_0\in K$ and a $\mathbb{C}$-basis $v$ of $V_{A_0}$. The matrix $B_A(v)$ has entries that are linear in $A$ (columns $v_k$ and $Av_k$). Since $\det B_{A_0}(v)\ne0$, there is a neighbourhood $U$ of $A_0$ in $M_n(\mathbb{R})$ on which $\det B_A(v)$ has the same sign and is nonzero. For $A\in U\cap K$, the vectors $v_k,Av_k$ ($k\le m$) form a real basis, so $v$ is a $\mathbb{C}$-basis of $V_A$ (its complex span contains all $v_k$ and $Av_k$). Hence $\varepsilon(A)=\operatorname{sign}\det B_A(v)=\varepsilon(A_0)$ on $U\cap K$. So $\varepsilon:K\to\{\pm1\}$ is continuous.

Step 4: the two pieces. In Step 1 we saw that for $A=PJ_0P^{-1}$ one may take $v_k=Pe_{2k-1}$ and then $B_A(v)=P$. Hence
$$\varepsilon(PJ_0P^{-1})=\operatorname{sign}\det P.$$
Therefore
$$K^+=\varepsilon^{-1}(+1)=\{PJ_0P^{-1}:P\in GL_n^+\},\qquad K^-=\varepsilon^{-1}(-1)=\{PJ_0P^{-1}:P\in GL_n^-\}.$$
Both are nonempty ($J_0\in K^+$; $DJ_0D^{-1}\in K^-$ for $D=\operatorname{diag}(-1,1,\dots,1)$). Each is the image of the connected set $GL_n^\pm$ under the continuous map $P\mapsto PJ_0P^{-1}$, hence connected. Since $\varepsilon$ is continuous, $K^+$ and $K^-$ are open and closed in $K=K^+\sqcup K^-$. Thus $K$ has exactly two components.

Check ($m=1$). For $n=2$ the construction gives the two sheets $\operatorname{Im}z=\pm\sqrt{1+|w|^2}$ of the problem on square roots of $-I$ in $M_2(\mathbb{R})$. A numerical test in $n=4$ with random $P$ confirms that $\operatorname{sign}\det B_A(v)=\operatorname{sign}\det P$ for random complex bases $v$ of $V_A$.

Remark. Another way to phrase it: $K$ is the conjugacy orbit $GL_n(\mathbb{R})/GL_m(\mathbb{C})$. The subgroup $GL_m(\mathbb{C})\subset GL_n^+$ is connected, so $\pi_0$ of the orbit equals $\pi_0(GL_n(\mathbb{R}))=\mathbb{Z}/2$. Also $A\mapsto-A$ changes $\varepsilon$ by $(-1)^m$: it swaps the two components iff $m$ is odd.

Lesson. To prove that a set has at least two components, produce a continuous invariant with two values; to prove at most two, exhibit it as the continuous image of a space with two components.