Questions about fixed point problem

Question

Consider a homeomorphism $f:[0,1]^n\rightarrow [0,1]^n$, n$\in\mathbb{N}$. Let $S:=\{x\in[0,1]^n:f(x)=x\}$

1. Can $S$ contain exactly two points?
2. If $S$ is not one point set, then does $S$ include a (nondegenerated) path?

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(we know $S\neq\emptyset$ by Brouwer fixed point theorem)

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I got to conjecture these from studying things related to fixed point problems, and these seem to be true to me. However, I don't have any idea to prove even 1, which is weaker than 2. (and I doubt that 2 is true)

Answer 1

$[0,1]^n$ is homeomorphic to the closed ball $B^n$ and that is the homeomorphic to the suspension of $B^{n-1}$. This is the product $[0,1]\times B^{n-1}$ with $0\times B^{n-1}$ identified to a point and $1\times B^{n-1}$ identified to (another) point. Consider $f:(t,x)\mapsto (t^2,x)$. The only fixed points on the suspension are the "poles" $0\times B^{n-1} $ and $1\times B^{n-1}$.