Do the homotopy groups of a space determine the homotopy groups of its suspension? I am trying to work out an answer to my latest question 
I know that there is not a known formula for the homotopy groups of the suspension in terms of the homotopy groups of the space being suspended (e.g. spheres), but I would like to know if the homotopy groups of $X$ determine the homotopy groups of $SX$. 
I suspect it is not the case, but no examples come to mind. 
 A: I have come up with a counterexample:
$S^2$ and $S^3 \times \mathbb{C}P ^\infty$ have the same homotopy groups, but their suspensions don’t since $\pi_7 (S^3)$ is torsion but $\pi_7(S(S^3 \times \mathbb{C}P ^\infty))$ contains an infinite element since $\pi_7(S^4)$ is not torsion. 
A: Since you yourself have already produced a counter example to what you were seeking I will provide here an example of the opposite of what you asked for. Namely I will observe the existence of an infinite family of spaces, all of which suspend to the same homotopy type. In particular $\pi_*\Sigma X$ cannot directly determine $\pi_*X$.
Consider the wedge $S^n\vee S^n$ with $n\geq 2$. Then it holds that 
$$\pi_{2n-1}(S^n\vee S^n)\cong \pi_{2n-1}S^2\oplus\pi_{2n-1}S^n\oplus \mathbb{Z}\{w\}$$
where the free factor is generated by the Whitehead product $w$. For each $k\in\mathbb{Z}$ let 
$$X_k=(S^n\vee S^n)\cup_{k\cdot w} e^{2n}$$
be the space obtained by attaching a $2n$-cell to $S^n\vee S^n$ with $k$ times $w$. Then $\Sigma w\simeq \ast$ since $w$ is in the kernel of the suspension homomorphism, and therefore for each $k$ you can find a homotopy equivalence
$$\Sigma X_k\simeq S^{n+1}\vee S^{n+1}\vee S^{2n+1}.$$
In particular
$$\pi_*(\Sigma X_k)\cong \pi_*(\Sigma X_l)$$
for any integers $k,l\in\mathbb{Z}$, so the homotopy groups of $\Sigma X_k$ are independent of the integer $k$. However it is not difficult to see that
$$\pi_*(X_k)\not\cong \pi_*(X_l)$$
if $k\neq\pm l$.
