The definition I have seen for an almost complex structure is the following
$$J:TM\to TM$$ which is linear fibre by fibre, such that $J^2 = -\text{Id}$, and such that $\pi(J(X_x)) = x$ where $\pi$ is the projection of $TM$ on $M$.
I see that this implies a map $$\tilde{J}:\mathfrak{X}(M)\to\mathfrak{X}(M)$$ such that $(\tilde{J})^2 = -\text{Id}$ and given by $\tilde{J}(X)(x) = J(X_x)$
My question is:
Given a linear map $\tilde{J}:\mathfrak{X}(M)\to\mathfrak{X}(M)$ such that $(\tilde{J})^2 = -\text{Id}$, can we reconstruct an almost complex structure on $M$?
Let $V$ be an infinite-dimensional vector space. By the axiom of choice, there is a basis $\{v_i\}_{i\in I}$ for $V$. As $I$ is infinite, we can find $A, B \subset I$ with $A\cap B = \emptyset$, $A\cup B = I$, and $|A| = |B|$. Let $\sigma : A \to B$ be a bijection. Then we can define a linear map $\tilde{J} : V \to V$ by setting $\tilde{J}(v_a) = v_{\sigma(a)}$ and $\tilde{J}(v_b) = -v_{\sigma^{-1}(b)}$; note that $\tilde{J}^2 = -\operatorname{id}_V$.
Now note that $\mathfrak{X}(M)$ is an infinite-dimensional vector space if $\dim M > 0$. If we could construct a $J$ from $\tilde{J}$, we would see that every positive-dimensional manifold admits an almost complex structure which is absurd.