Show that any almost complex structure is induced by at most one complex structure.
Suppose $X$ is a complex manifold of dimension $n$ and $M$ is the underlying $2n$ dimensional real manifold.
If $J_1$ and $J_2$ are two almost complex structures induced by the same complex structure on $X$ I would want to show that $J_1 = J_2$.
I know that any almost complex structure induced by an complex structure is integrable so for both $J_1$ and $J_2$ we have that the Nijenhuis tensor $$N(X,Y)=[X,Y]+J([JX,Y] + [X,JY]) - [JX,JY]$$ vanishes. I'm wondering on whether this is enough to conclude that $J_1=J_2$?
To address the problem in the title, rather than the problem you set up in the subsequent text: let $\mathcal{X}_1$ and $\mathcal{X}_2$ denote two complex structures on an underlying manifold $X$. There are corresponding bundle maps $J_1,J_2:TX\to TX$. Suppose that $J_1=J_2$, or (more generally) that $f^*J_1=J_2$ for some diffeomorphism $f:X\to X$. Then we must show that $\mathcal{X}_1\cong\mathcal{X}_2$, as complex manifolds. This amounts to checking that $f$ (which in the simplest case is $f=\text{id}$) is a biholomorphism. Now prove the following:
Lemma. Let $f:\mathcal{X}\to\mathcal{Y}$ be a smooth map between complex manifolds with induced almost complex structures $J_X$ and $J_Y$. Then $f$ is holomorphic if and only if $df\circ J_X=J_Y\circ df$.
Using the lemma, you can now show that the assumption that $f^*J_1=J_2$ implies that $f$ must be a biholomorphism, and therefore $\mathcal{X}_1\cong\mathcal{X}_2$.