Here is an example of unnaturally isomorphic functors for which there does not exist any natural transformation between them.

Let $\mathcal{C} = \mathbb{N}^{\mathrm{op}}$, $\mathcal{D} = \mathcal{Ab}$, and consider $F, G : \mathcal{C} \rightarrow \mathcal{D}$ defined by $$F(n) = G(n) = \mathbb{Z} \quad\text{for all } n \in \mathbb{N},$$ $$F(m \le n)(x) = 2^{n-m} x,\quad G(m \le n)(x) = 3^{n-m} x \quad\text{for all } x \in \mathbb{Z}.$$

Suppose $\eta = \{ \eta_n : F(n) \rightarrow G(n) \}_{n \in \mathbb{N}}$ is a natural transformation. Then for any $n$ we have that $\eta_0 (2^n x) = 3^n \eta_n (x)$, so $2^n \eta_0 (x) = 3^n \eta_n(x)$. But then $3^n \mid \eta_0 (x)$ for all $n$, which is clearly absurd.

Here is an example of unnaturally isomorphic functors for which there does not exist any non-trivial natural transformation between them.

Let $\mathcal{C} = \mathbb{N}^{\mathrm{op}}$, $\mathcal{D} = \mathcal{Ab}$, and consider $F, G : \mathcal{C} \rightarrow \mathcal{D}$ defined by $$F(n) = G(n) = \mathbb{Z} \quad\text{for all } n \in \mathbb{N},$$ $$F(m \le n)(x) = 2^{n-m} x,\quad G(m \le n)(x) = 3^{n-m} x \quad\text{for all } x \in \mathbb{Z}.$$

Suppose $\eta = \{ \eta_n : F(n) \rightarrow G(n) \}_{n \in \mathbb{N}}$ is a natural transformation. Then for any $n$ we have that $\eta_0 (2^n x) = 3^n \eta_n (x)$, so $2^n \eta_0 (x) = 3^n \eta_n(x)$. But then $3^n \mid \eta_0 (x)$ for all $n$, which implies that $\eta_0(x) = 0$, and so $\eta = 0$.