Let $X$ be a topological space. Let $\Sigma$ be suspension.
Does $H^n(X;\mathbb{Z})\cong H^{n+1}(\Sigma X;\mathbb{Z})$ isomorphic or not?
Does $H^n(X;\mathbb{Z}_2)\cong H^{n+1}(\Sigma X;\mathbb{Z}_2)$ isomorphic or not?
Does $H^n(X;\mathbb{Z}_p)\cong H^{n+1}(\Sigma X;\mathbb{Z}_p)$ isomorphic for odd prime $p$ or not?
I tried to use long exact sequence of cohomology induced from $S^1\vee X\to S^1\times X\to \Sigma X$ and apply Kunneth formula on $H^*(S^1\times X;\mathbb{Z}_2)$, but I cannot work out.
Here's the argument given in the comment as an answer. Coefficients should be understood to be $\Bbb Z$.
Consider the pair $(CX,X)$, where $CX = X \times [0,1]/\sim$ , where $(x,1) \sim (x',1)$ for all $x,' \in X$, and $X$ is considered as the subspace $X \times \{0\}$ of $CX$. Then $CX/X \cong \Sigma X$.
Now we have a long exact sequence in reduced homology $$\dots \to H_{n+1}(CX,X) \to H_n(X) \to H_n(CX) \to H_n(CX,X) \to \dots$$
Now $CX$ is contractible so has trivial homology, so by exactness $H_{n+1}(CX,X) \cong H_n(X)$. Because $X$ is a closed subset of $CX$, we have that $H_{n+1}(CX,X) \cong H_{n+1}(CX/X) \cong H_{n+1}(\Sigma X)$, hence $H_{n+1}(\Sigma X) \cong H_n(X)$. Finally, thanks to the universal coefficient theorem for cohomology (Theorem 3.2 in https://pi.math.cornell.edu/~hatcher/AT/AT.pdf) we get the desired results.