Suppose that a path-connected CW complex $X$ is homotopy equivalent to its own suspension $SX$. Is it true that $X$ must be contractible?
I have tried to show that since $X \times I \simeq SX = X \times I /\! \sim$, the projection $X \times I \to X \times \lbrace 0 \rbrace$ induces a projection $X \times I /\! \sim\; \to \lbrace x_0 \rbrace \times \lbrace 0 \rbrace$ such that $X \times \lbrace 0 \rbrace \simeq \lbrace x_0 \rbrace \times \lbrace 0 \rbrace$ as desired, but this does not seem to be successful or true in general.
Edit: Following freakish's comment, by the Freudenthal suspension theorem and Whitehead's theorem, this is true. Might a more elementary proof be possible? (I don't think so, but I'd be pleasantly surprised if there was one.)
remark: you don't need any of this.
Making a few assumptions rather than abstract isomorphisms might help.
Suppose that $X$ is connected and a cw-complex. Also suppose that the natural map $X \to CX/X=\Sigma X$ induces a weak equivalence.
Instead of freudenthal suspension (which is sort of hardcore) we can use Van-Kampen's theorem to show that $X$ is simply connected.
Then the Hurewicz theorem improves Whitehead's theorem so that a homology equivalence is enough.
from the long exact sequence we get an exact sequence $ \cdots \to \tilde{H_n}(X) \to \tilde{H}_n(CX) \to \tilde{H}_n(\Sigma X) \to \cdots$
Now the composite of the first two maps is assumed to be an isomorphism but factors through zero, so in fact $\tilde{H_n}(X)=0$ and this isomorphism is induced by the inclusion $X \to CX$.
This suffices for your question on contractibility of $S^\infty$, but there are much easier proofs using left and right shift operators.