Suppose $J :TM \to TM$ is an almost complex structure on $M$, and let $\varphi : TM \to TM$ be a bundle automorphism. Set $J_{\varphi} = \varphi\circ J\circ \varphi^{-1}$, then
\begin{align*}
J_{\varphi}\circ J_{\varphi} &= \varphi\circ J\circ \varphi^{-1}\circ\varphi\circ J\circ \varphi^{-1}\\
&= \varphi\circ J\circ J\circ \varphi^{-1}\\
&= \varphi\circ (-\operatorname{id}_{TM})\circ\varphi^{-1}\\
&= -\varphi\circ\operatorname{id}_{TM}\circ\varphi^{-1}\\
&= -\operatorname{id}_{TM}
\end{align*}
where we have used the fact that $\varphi$ is linear on fibres in the penultimate equality.
Therefore, if $M$ admits an almost complex structure $J$, every bundle automorphism $\varphi$ gives rise to another almost complex structure $J_{\varphi}$, but not every almost complex structure on $M$ arises in this way; for example, $-J$.
Note, the previous paragraph is nothing more than a global version of the following linear algebra statement: if $J$ is a matrix which squares to $-I_n$, then any matrix which is similar to $J$ also squares to $-I_n$, but not every such matrix is similar to $J$; for example, $-J$.
Best Answer
The classic example of an almost complex manifold that is not a complex manifold is the six-sphere $S^6$. Consider $S^6$ in $\mathbb{R}^7 = \mathrm{im}\,\mathbb{O}$ as the set of unit norm imaginary octonions. The almost complex structure on $S^6$ is defined by $J_p v = p \times v$, where $p\in S^6$ and $v\in T_p S^6$ and $\times$ stands for the vector product on $\mathbb{R}^7$. This almost complex structure cannot be induced by a complex atlas on $S^6$ because the Nijenhuis tensor $N_J$ doesn't vanish (cf. the Nirenberg Newlander theorem).
A result of Borel-Serre states that the only spheres endowed with an almost complex structure are $S^2$ and $S^6$. (The one on $S^2$ is a complex structure.) Up to now, it is not known whether there exists a complex structure on $S^6$.
As a reference I mention the Wikipedia page on almost complex manifolds.