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1. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". Then we have $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for every $X \in C$, but $\mathrm{Sym} \not\cong \mathrm{Ord}$ (in fact between these functors there is no natural transformation at all).

2. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

3. If $C$ denotes the category of finite abelian groups, then $\mathrm{Tor}_1^{\mathbb{Z}}$ and $\otimes_{\mathbb{Z}} : C \times C \to C$ are pointwise isomorphic (since $\mathrm{Tor}_1(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m) \cong \mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Z}/m$), but they are not isomorphic (for example since $\mathrm{Tor}_1^{\mathbb{Z}}$ is not right exact in the second or first variable).

If $F,G : C \to D$ are functors such that $F(x) \cong G(x)$ for every $x \in C$, I would call $F,G$ "pointwise isomorphic". You ask for examples of non-isomorphic functors which are pointwise isomorphic. There are plenty natural examples.

1. Consider the interval category $I=\{0 \to 1\}$. The category of functors $I \to C$ is isomorphic to the category of morphisms in $C$. Of course for most $C$ there are non-isomorphic morphisms in $C$ whose domain and codomain are isomorphic or even equal. For example take the identity and a constant map on a nontrivial set or space.

2. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". These functors are pointwise isomorphic, but not isomorphic (in fact between these functors there is no natural transformation at all). Actually this example (when restricted to sets of a given size) can be seen as a special case of the next one.

3. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

4. If $C$ denotes the category of finite abelian groups, then $\mathrm{Tor}_1^{\mathbb{Z}}$ and $\otimes_{\mathbb{Z}} : C \times C \to C$ are pointwise isomorphic (since $\mathrm{Tor}_1(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m) \cong \mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Z}/m$), but they are not isomorphic (for example since $\mathrm{Tor}_1^{\mathbb{Z}}$ is not right exact in the second or first variable).

1. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". Then we have $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for every $X \in C$, but $\mathrm{Sym} \not\cong \mathrm{Ord}$ (in fact between these functors there is no natural transformation at all).

2. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

3. If $C$ denotes the category of finite abelian groups, then $\mathrm{Tor}_1^{\mathbb{Z}}$ and $\otimes_{\mathbb{Z}} : C \times C \to C$ are pointwise isomorphic (since $\mathrm{Tor}_1(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m) \cong \mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Z}/m$), but they are not isomorphic.

1. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". Then we have $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for every $X \in C$, but $\mathrm{Sym} \not\cong \mathrm{Ord}$ (in fact between these functors there is no natural transformation at all).

2. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

3. If $C$ denotes the category of finite abelian groups, then $\mathrm{Tor}_1^{\mathbb{Z}}$ and $\otimes_{\mathbb{Z}} : C \times C \to C$ are pointwise isomorphic (since $\mathrm{Tor}_1(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m) \cong \mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Z}/m$), but they are not isomorphic (for example since $\mathrm{Tor}_1^{\mathbb{Z}}$ is not right exact in the second or first variable).

1. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". Then we have $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for every $X \in C$, but $\mathrm{Sym} \not\cong \mathrm{Ord}$ (in fact between these functors there is no natural transformation at all).

2. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

1. Let $C$ be the category of finite sets with bijections as morphisms. Then we have the functor $\mathrm{Sym} : C \to C$ which maps every set to its set of permutations, and the functor $\mathrm{Ord} : C \to C$ which maps every set to its set of total orderings; the action on morphisms is "conjugation". Then we have $\mathrm{Sym}(X) \cong \mathrm{Ord}(X)$ for every $X \in C$, but $\mathrm{Sym} \not\cong \mathrm{Ord}$ (in fact between these functors there is no natural transformation at all).

2. Let $G$ be a group (or monoid), considered as a category with one object $\star$. Then a functor $G \to \mathsf{Set}$ is the same as a $G$-set. In fact, the category of $G$-sets is isomorphic to the category of functors $G \to \mathsf{Set}$. The value at $\star$ is the underlying set. Of course for $G \neq 1$ there are non-isomorphic $G$-sets whose underlying sets are isomorphic (for example the underlying set of $G$ with the regular action and with the trivial action of $G$).

3. If $C$ denotes the category of finite abelian groups, then $\mathrm{Tor}_1^{\mathbb{Z}}$ and $\otimes_{\mathbb{Z}} : C \times C \to C$ are pointwise isomorphic (since $\mathrm{Tor}_1(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m) \cong \mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Z}/m$), but they are not isomorphic.