$(\phi: A \rightarrow A\wedge \phi \text{ strictly increasing})\rightarrow(\forall a \in A \phi(a)\geq a)\implies A \text{ well ordered}$

Question

Let $(A,<)$ be a total order.

Is the following true?

$\forall \phi: A \rightarrow A((\phi \text{ strictly increasing})\rightarrow(\forall a \in A \phi(a)\geq a))\implies A \text{ well ordered}$

We know that the opposite implication is true, but I can't prove or disprove the above assertion.

I think that the thesis is false but I was not able to find a counterexample.