The claim (which I can't find anywhere in the answers to the linked question) is incorrect. A constant function can't be one-way. To see why, let's recall the definition of a one-way function.
A function $f : \{0,1\}^* \to \{0,1\}^*$ is one-way, if
- There exists a polynomial time algorithm $M_f$ such that $M_f(x) = f(x)$ for all $x\in\{0,1\}^*$.
- For every PPT algorithm $\mathcal{A}$ there exists a negligible function $\mathsf{negl}$ such that for all $\kappa\in\mathbb{N}$ it holds that
$$\Pr[x\gets\{0,1\}^\kappa, y:=f(x)\;:\;f(\mathcal{A}(1^\kappa,y))=y ] \leq \mathsf{negl}(\kappa)$$
However, for any constant function is is easy to specify a PPT algorithm $\mathcal{A}$ for which
$$\Pr_{x\gets\{0,1\}^\kappa}\bigl[f\bigl(\mathcal{A}(1^\kappa,f(x))\bigr)=f(x)\bigr] = 1$$
for all $\kappa\in\mathbb{N}$.
E.g., we can define $\mathcal{A}$ as the algorithm that always outputs $1^\kappa$. I.e., for all $x\in\{0,1\}^\kappa$ we have $f\bigl(\mathcal{A}(1^\kappa,f(x))\bigr) = f(1^\kappa)$ and since the function $f$ is constant, it holds for all $x\in\{0,1\}^\kappa$ that $f(1^\kappa) = f(x)$. Thus $\mathcal{A}$ breaks the one-wayness of $f$ with probability $1$ and $f$ is not one-way.
