Natural isomorphism $\tilde H_i(X) \xrightarrow{\cong} \tilde H_{i+1}(\Sigma X)$ where $\Sigma X$ is the suspension of $X$. Define $\Sigma X$ to be the quotient space of $[-1,1]\times X$ obtained by identifying ${0}\times X$ and ${1}\times X$ to two points respectively. For any homology theory (satisfying Eilenberg-Steenrod axioms), I am able to find an isomorphism $\tilde H_i(X) \rightarrow \tilde H_{i+1}(\Sigma X)$ as follows:
Denote $C_+ X=[0,1]\times X /\tilde{}$ and $C_- X=[-1,0]\times X /\tilde{}$, then we know both of them are contractible and $\Sigma X = C_+X \cup C_- X$. First we consider the reduced homology sequence for the pair $(\Sigma X, C_+ X)$:
$$
0=\tilde H_i(C_+ X) \to \tilde H_i(\Sigma X) \to H_i(\Sigma X,C_+ X) \to \tilde H_{i-1}(C_+ X)=0
$$
Hence we know $$\tilde H_i(\Sigma X)  \to^{\cong} H_i(\Sigma X,C_+ X)$$ is an isomorphism. Second, by considering the reduced homology sequence for the pair $(C_- X, X)$(where X is identified with the quotient image of $\{0\}\times X$), we can similarly get $$
H_i(C_-X,X)\to^{\cong} \tilde H_{i-1}(X)
$$
Finally, using excision axiom and homotopy axiom we can show that 
$$
H_i(C_-X,X) \cong H_i(\Sigma X,C_+ X)
$$

Nevertheless, I have no idea how to show this isomorphism is also
  "natural".  Here "natural" means that, if we denote this isomorphism
  by $\Phi: \tilde H_i(X) \to \tilde H_{i+1}(\Sigma X)$, then, for any
  map $f:X\to Y$ and its suspension $\Sigma f: \Sigma X \to \Sigma Y$,
  $\Phi f_* = (\Sigma f)_* \Phi$.

 A: All the constructions that you used to define the isomorphism are natural/functorial:


*

*Given a map $X \to Y$, you have a natural map that respect inclusions, which gives a starting point for all the applications of naturality to come:
$$(\Sigma X, C_+ X, C_- X, X \times \{0\}) \to (\Sigma Y, C_+ Y, C_- Y, Y \times \{0\});$$

*The long exact sequence of a pair is natural, hence by using the natural map $(\Sigma X, C_+ X) \to (\Sigma Y, C_+ Y)$, the isomorphism $\tilde{H}_i(\Sigma X) \to \tilde{H}_i(\Sigma X, C_+ X)$ is natural in $X$;

*Excision is natural, hence the excision isomorphism $\tilde{H}_i(C_- X, X) \to \tilde{H}_i(\Sigma X, C_+ X)$ is natural in $X$;

*Finally the long exact sequence is still natural, hence the isomorphism $\tilde{H}_i(C_- X, X) \to \tilde{H}_{i-1}(X)$ is natural in $X$.


In conclusion, each subsquare in the following diagram is commutative, hence the "outer" rectangle (by inverting the horizontal arrows that go the wrong way, the composite of the whole thing is the suspension isomorphism) is commutative:
$$\require{AMScd}
\begin{CD}
\tilde{H}_i(\Sigma X) @>{\cong}>> \tilde{H}_i(\Sigma X, C_+ X) @<{\cong}<< \tilde{H}_i(C_- X, X) @>{\cong}>> \tilde{H}_{i-1}(X) \\
@VVV @VVV @VVV @VVV \\
\tilde{H}_i(\Sigma Y) @>{\cong}>> \tilde{H}_i(\Sigma Y, C_+ Y) @<{\cong}<< \tilde{H}_i(C_- Y, Y) @>{\cong}>> \tilde{H}_{i-1}(Y)
\end{CD}$$
tl;dr The composition of two functors is a functor.
