No. Here is a construction.

It is not hard to see that there is an orientation-preserving isometry $L:\mathbb{H}\to\mathbb{H}$ (where $\mathbb{H}\simeq\mathbb{R}^4$ is the ring of quaternions) such that $L^2=1$, namely, $$ L(x) = \tfrac12 (j+k)\,x\,(j+k). $$ This isometry satisfies $L(jx)=kL(x)$.

Now let $\Lambda\subset\mathbb{H}$ be a lattice that is preserved by $L$ (generated by lattices in the two $2$-dimensional eigenspaces of $L$) and let $M = \mathbb{H}/\Lambda$. Then $L$ induces an orientation-preserving isometric involution $\phi$ of $M$.

Moreover, left multliplication by $j$ (respectively, $k$) defines a translation-invariant orthogonal complex structure $J$ (respectively, $K$) on $\mathbb{H}$ that descends to $M$, and, by the equation $L(jx)=kL(x)$, we have $\phi^*(J) = K$.

Note that, with the quotient flat metric $g$ on $M$, we have that $(M,J,g)$ and $(M,K,g)$ are Kähler manifolds and $\phi$ is an orientation-preserving isometric involution of $g$.

For any tangent vector $v$, the vectors $v$, $Jv$, and $Kv$ are linearly independent, hence there are no $2$-dimensional subspaces of the tangent space at any point that are both $J$ and $K$ invariant. Thus, the only possibility would be to take $W_i = T_pM$, but $J\not=\pm K$ on $T_pM$ for any $p\in M$.

No. Here is a construction.

It is not hard to see that there is an orientation-preserving isometry $L:\mathbb{H}\to\mathbb{H}$ (where $\mathbb{H}\simeq\mathbb{R}^4$ is the ring of quaternions) such that $L^2=1$, namely, $$ L(x) = \tfrac12 (j+k)\,x\,(j+k). $$ This isometry satisfies $L(jx)=kL(x)$.

Now let $\Lambda\subset\mathbb{H}$ be a lattice that is preserved by $L$ (for example, it could be generated by lattices in the two $2$-dimensional eigenspaces of $L$), and let $M = \mathbb{H}/\Lambda$. Then $L$ induces an orientation-preserving isometric involution $\phi$ of $M$.

Moreover, left multliplication by $j$ (respectively, $k$) defines a translation-invariant orthogonal complex structure $J$ (respectively, $K$) on $\mathbb{H}$ that descends to $M$, and, by the equation $L(jx)=kL(x)$, we have $\phi^*(J) = K$.

Note that, with the quotient flat metric $g$ on $M$, we have that $(M,J,g)$ and $(M,K,g)$ are Kähler manifolds and $\phi$ is an orientation-preserving isometric involution of $g$.

For any tangent vector $v$, the vectors $v$, $Jv$, and $Kv$ are linearly independent, hence there are no $2$-dimensional subspaces of the tangent space at any point that are both $J$ and $K$ invariant. Thus, the only possibility would be to take $W_i = T_pM$, but $J\not=\pm K$ on $T_pM$ for any $p\in M$.