**Edited (after R. Bryant comment)**


Let $(M,\cal J,g)$ be a almost Hermitian manifold (*not necessary integrable*). i.e., ${\cal J}^2=-I$ and $g({\cal J} X,{\cal J} Y)=g(X,Y)$. Suppose that $\{X\_i,{\cal J}X\_i\}$ be **any** local orthonormal ${\cal J}$-frame and the following relation hold for $i\neq j$ 
$$g(Q{\cal J}X\_i,{\cal J}X\_i)=g(QX\_j,X\_j),\quad K(X\_i,X\_j)=K(X\_i,{\cal J}X\_i);$$
where $Q$ and $K$ are Ricci operator and sectional curvature respectively.
Then 



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


Your advice or suggestions will be much appreciated and welcomed.



Almost Complex manifolds of constant curvature
Well, first of all, your conditions are vacuous if the dimension of $M$ is $2$, and the conclusion in that case is false. Thus, you must also assume, in order to get the conclusion, that the dimension of $M$ is $2n>2$.


It turns out that the answer is 'yes' if $M$ has dimension $4$, but my proof is not particularly simple. I suspect that the answer is 'yes' for all dimensions greater $2n>4$, but I have not checked this yet.


Meanwhile, I think it is worth pointing out that, when $M$ has dimension $2n>2$, your condition
$$
K(X\_i,X\_j) = K(X\_i,\mathcal{J}X\_i)\tag 1
$$
for all local orthonormal $\mathcal{J}$-frames implies that $(M,g)$ is Einstein, 
so, in particular, the relation
$$
g(Q\mathcal{J}X\_i,\mathcal{J}X\_i) = g(QX\_j,X\_j) \tag 2
$$
is an automatic consequence.


To see this, fix a point $x\in M$ and let $V = T\_xM$, endowed with its inner product $g\_x$ and complex structure $\mathcal{J}\_x$. Let $B\_x\subset V\times V$ denote the set of pairs $(v,w)$ such that $v$ and $w$ are unit vectors that satisfy $g\_x(v,w) = g\_x(\mathcal{J}\_xv,w) = 0$. This $B\_x$ is a smooth, connected manifold of dimension $4n{-}4$. It has a projection $\ell:B\_x\to\mathbb{P}(V)\simeq \mathbb{CP}^{n-1}$ defined by $\ell(v,w) = \mathbb{C}v$, and another projection $p:B\_x\to\mathrm{Gr}^+\_2(V)$, the space of oriented unit $2$-planes in $V$, defined by $p(v,w) = v\wedge w$ whose image is the smooth hypersurface $L\_x\subset\mathrm{Gr}^+\_2(V)$ of dimension $4n{-}5$ consisting of the $2$-planes on which the canonical $(1,1)$-form $\omega$ vanishes, where $\omega(v, w) = g\_x(\mathcal{J}\_xv,w)$. 


Consider the function $K:B\_x\to\mathbb{R}$ defined by the sectional curvature $K(v\wedge w)$. This function is constant on the fibers of $p$ by definition, but it is also constant on the fibers of $\ell$ by (1). Now, it is easy to see that these two constancies imply that $K$ is constant on all of $B\_x$. (One way to see this is to note that any two points of $B\_x$ can be connected by a sequentially connected chain of $p$-fibers and $\ell$-fibers.) Thus, there is a smooth function $\sigma:M\to\mathbb{R}$ such that $K(v\wedge w) = \sigma(x)$ for all $(v,w)\in B\_x$, and, moreover, $K(v\wedge \mathcal{J}\_xv) = \sigma(x)$ for all $v\in V = T\_xM$.


Now, if $v$ is any unit vector in $T\_xM$, then there exists a orthonormal $\mathcal{J}$-frame $(X\_i,\mathcal{J}X\_i)$ on a neighborhood of $x$ such that $X\_1(x) = v$. Consequently,
$$
\mathrm{Ric}(v,v) = K\bigl(v\wedge \mathcal{J}\_xv\bigr) + \sum\_{i=2}^nK\bigl(v\wedge X\_i(x)\bigr)+ K\bigl(v\wedge \mathcal{J}\_xX\_i(x)\bigr)
= (2n{-}1)\,\sigma(x).
$$
In particular, it follows, since $v$ is any unit vector in $T\_xM$, that $\mathrm{Ric} = (2n{-}1)\sigma\, g$. Since the dimension of $M$ is greater than $2$, Schur's Lemma implies that $\sigma$ must be constant on $M$, i.e., that $g$ is Einstein, which, of course implies (2).


Now, it is well known that if the sectional curvature is constant, i.e, if $K(v\wedge w) = \sigma$ for *all* $v\wedge w\in\mathrm{Gr}^+\_2(TM)$, then $(M,g)$ has constant curvature. However, we do not yet know this property. We only know that $K$ is fiberwise constant on the hypersurface bundle $L\subset \mathrm{Gr}^+\_2(TM)$ unioned with the smaller subbundle $\mathbb{P}(TM)\subset \mathrm{Gr}^+\_2(TM)$ consisting of the $2$-planes that are complex lines (in either orientation) and that the scalar curvature function is constant. This, by itself, is not sufficient to conclude that the Riemann curvature tensor has constant sectional curvature.


For example, when $2n=4$ (the first nontrivial case), when given a $(M,g,\mathcal{J})$, i.e., a $\mathrm{U}(2)$-structure on $M$, the space of potential Riemann curvature tensors for which the sectional curvature $K$ is a given constant $\sigma$ on the union of $L$ and $\mathbb{P}(TM)$ are the sections of an affine bundle of rank $2$ over $M$. One can set up the structure equations for such a $\mathrm{U}(2)$-structure over $M$ and, by applying the Bianchi identities a few times and differentiating and combining results, conclude that any such structure (even a local one) must, in fact, have constant sectional curvature. Thus, when $2n=4$, the answer to your question is 'yes'.


However, when $2n>4$, things are not so clear. The first task is to describe the bundle of possible Riemann curvature tensors for $\mathrm{U}(n)$-structures on $M^{2n}$ with the property that the sectional curvature $K$ is constant on $L\cup \mathbb{P}(TM)\subset \mathrm{Gr}^+\_2(TM)$. Again, it's an affine bundle, but I don't yet know its rank. It may be that, when $2n>4$, the constant sectional curvature tensor is the only one that satisfies (1), but I have not shown this yet. If that is the case, then the answer will be 'yes' in those cases as well, with a much easier proof than the case $2n=4$.


I am interested to know what led you to consider the conditions (1) and (2).


