($\Rightarrow$) I'll just demonstrate that $\kappa \rightarrow ( \kappa )^1_{\mathrm{cf} (\kappa)}$ fails.
Denoting $\mathrm{cf} ( \kappa)$ by $m$, let $\langle \alpha_i \rangle_{i < m}$ be a strictly increasing sequence cofinal in $\kappa$. Define $f : \kappa \to m$ by setting $f(\alpha)$ to be the least $i$ such that $\alpha \leq \alpha_i$. It is easy to see that any $f$-homogeneous set must be bounded, and, since $\kappa$ is a cardinal, bounded subsets have cardinality strictly less than $\kappa$.
($\Leftarrow$) Assume first that $\kappa > \lambda$ (and \kappa > m), then given $f : \kappa \to m$, consider $\{ f^{-1} \{i\} : i < m \}$. If none of these sets has size at least $\lambda$, then $\kappa = | \kappa | = \left| \bigcup_{i < m} f^{-1} \{ i \} \right| \leq \sum_{i < m} \left| f^{-1} \{ i \} \right| \leq m \cdot \lambda < \kappa,$ a contradiction!
If $\kappa = \lambda$ and $\mathrm{cf} (\kappa) > m$, given $f : \kappa \to m$, consider $\{ f^{-1} \{i\} : i < m \}$. If none of these sets has size $\kappa$, I claim that there is a $\mu < \kappa$ such that $\left| f^{-1} \{ i \} \right| \leq \mu$ for all $i < m$.
- If $\kappa$ is a successor cardinal, then $\kappa = \mu^+$ for some $\mu$, and it easily follows from the above that $\left| f^{-1} \{ i \} \right| \leq \mu$ for all $i < m$.
- If $\kappa$ is a limit cardinal, then the family of all cardinals $\mu < \kappa$ is cofinal in $\kappa$. As $\left\{ \left| f^{-1} \{ i \} \right| : i < m \right\}$ is a family of $\,\leq m < \mathrm{cf} (\kappa)$ cardinals less than $\kappa$, it follows that it cannot be cofinal in $\kappa$, and so there must be a $\mu < \kappa$ with $\left| f^{-1} \{ i \} \right| \leq \mu$ for all $i < m$.
Taking such a $\mu$, it follows that $\kappa = | \kappa | = \left| \bigcup_{i < m} f^{-1} \{ i \} \right| \leq \sum_{i < m} \left| f^{-1} \{ i \} \right| \leq m \cdot \mu < \kappa,$ a contradiction!