Almost Hermitian manifolds of constant curvature Let $(M,g)$ be a $2n$-dimensional almost Hermitian manifold ($n\geq 2$) with a almost complex structure $\cal J$ (not necessary integrable). i.e.,
 $${\cal J}^2=-I,\quad\qquad g({\cal J} X,{\cal J} Y)=g(X,Y).$$
Suppose that $\{X_i,{\cal J}X_i\}$ be any local orthonormal ${\cal J}$-frame and the following relations hold for $i\neq j$ 
$$K(X_i,X_j)-K(X_j,{\cal J}X_j)=\frac{1}{2n-2}\{\rho(X_i,X_i)-\rho({\cal J}X_j,{\cal J}X_j)\}\tag{1}$$
$$K(X_i,X_j)-K(X_i,{\cal J}X_j)=\frac{1}{2n-2}\{\rho(X_j,X_j)-\rho({\cal J}X_j,{\cal J}X_j)\}\tag{2}$$
where $\rho$ and $K$ are Ricci curvature tensor and sectional curvature respectively.
Then 

Can be deduce that $(M,\cal J,g)$ is of constant curvature?

This question comes from the study of conformally flat almost Hermitian manifolds.
Your suggestions will be appreciated.
 A: The answer to the question as asked is 'No', at least when $2n=4$.  In that case, the conditions (1) and (2) only involve the Weyl curvature of the underlying metric $g$, so, in particular, when $g$ is conformally flat, then the conditions are satisfied, and there are many conformally flat metrics in dimension $4$ that do not have constant curvature.  
To construct an example of $(g,\mathcal{J})$ that does not have constant curvature just choose any conformally flat metric $g$ in dimension $4$ that does not have constant sectional curvature and any $g$-compatible almost complex structure $\mathcal{J}$.
Since the Ricci curvature is not constrained, there are probably examples that are not conformally flat, but they do, at least, have to be self-dual, i.e., $W_-$ must vanish (which is implied by (1) and (2)).  Meanwhile, the conditions (1) and (2) imply that $W_+$ must take values in a certain rank $2$ subbundle of the self-dual Weyl curvatures that is defined using $\mathcal{J}$:  If $\omega$ is the the (self-dual) $2$-form defined so that $\omega(v,w) = g(\mathcal{J}_xv,w)$ for all $v,w\in T_xM$, then the self-dual $2$-forms can be split as $\Lambda^2_+(T^*M) = \mathbb{R}\omega\oplus E$ where $E$ is a bundle of rank $2$.  Then (1) and (2) are equivalent to the requirement that the Weyl curvature be self-dual and take values in the rank $2$ subbundle $\omega\circ E\subset S^2\bigl(\Lambda^2_+(T^*M)\bigr)$.
