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As has been pointed out, the putative converse to the Contraction Mapping Theorem suggested in the question is not true. But there is a result which may reasonably be viewed as the converse of CMT.

Theorem (Bessaga, 1959): Let $X$ a set and $f: X \rightarrow X$ a function such that for all $n \in \mathbb{Z}^+$, the iterate $f^n$ has a unique fixed point. Then there exists a complete metric on $X$ with respect to which $f$ is a contraction mapping (for any preassigned constant $c \in (0,1)$).

[Addendum: I just looked at Elekes' paper and saw that it cites this result of Bessaga and says that it is seemingly the earlist earliest converse. So I guess this post is not exactly exciting news. Oh well.]

In later correspondence with him I pointed out the following result, which is now included in his writeup:

Theorem: For a function $f: X \rightarrow X$, the following are equivalent:
(i) Every iterate $f^n$ has at most one fixed point.
(ii) There is a metric (not necessarily complete!) with respect to which $f$ is a contraction.

This is not an earth-shattering result but it has a nice, crisp statement and afterwards I decided that it was too good to be true that I was the first to think of it. And I was right -- after a quick internet search I found the result in a published paper. (Unfortunately I didn't take note of the reference. Sorry, K.)

2 added 214 characters in body

As has been pointed out, the putative converse to the Contraction Mapping Theorem suggested in the question is not true. But there is a result which may reasonably be viewed as the converse of CMT.

Theorem (Bessaga, 1959): Let $X$ a set and $f: X \rightarrow X$ a function such that for all $n \in \mathbb{Z}^+$, the iterate $f^n$ has a unique fixed point. Then there exists a complete metric on $X$ with respect to which $f$ is a contraction mapping (for any preassigned constant $c \in (0,1)$).

[Addendum: I just looked at Elekes' paper and saw that it cites this result of Bessaga and says that it is seemingly the earlist converse. So I guess this post is not exactly exciting news. Oh well.]

In later correspondence with him I pointed out the following result, which is now included in his writeup:

Theorem: For a function $f: X \rightarrow X$, the following are equivalent:
(i) Every iterate $f^n$ has at most one fixed point.
(ii) There is a metric (not necessarily complete!) with respect to which $f$ is a contraction.

This is not an earth-shattering result but it has a nice, crisp statement and afterwards I decided that it was too good to be true that I was the first to think of it. And I was right -- after a quick internet search I found the result in a published paper. (Unfortunately I didn't take note of the reference. Sorry, K.)

1

As has been pointed out, the putative converse to the Contraction Mapping Theorem suggested in the question is not true. But there is a result which may reasonably be viewed as the converse of CMT.

Theorem (Bessaga, 1959): Let $X$ a set and $f: X \rightarrow X$ a function such that for all $n \in \mathbb{Z}^+$, the iterate $f^n$ has a unique fixed point. Then there exists a complete metric on $X$ with respect to which $f$ is a contraction mapping (for any preassigned constant $c \in (0,1)$).

Theorem: For a function $f: X \rightarrow X$, the following are equivalent:
(i) Every iterate $f^n$ has at most one fixed point.
(ii) There is a metric (not necessarily complete!) with respect to which $f$ is a contraction.