There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.

As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.

Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.

On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)

From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$.

View $G=C\rtimes D$ as a subgroup of $\text{GL}(5,3)$. From Jordan's normal form theorem, we see that $\text{GL}(5,3)$ contains two conjugacy classes of elements of order $9$. Let $U$ be the group of upper triangular matrices of $\text{GL}(5,3)$. Consider the two cases of $C$ (corresponding to the Jordan block sizes $3+1+1$ and $3+2$). In both cases, one computes that the exponent of $N_U(C)/C$ is $3$, so there is no room for the cyclic group $D$ of order $9$.

There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.

As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.

Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.

On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)

From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$.

View $G=C\rtimes D$ as a subgroup of $\text{GL}(5,3)$. From Jordan's normal form theorem, we see that $\text{GL}(5,3)$ contains two conjugacy classes of elements of order $9$. Let $U$ be the group of upper triangular matrices of $\text{GL}(5,3)$. Consider the two cases of $C$ (corresponding to the Jordan block sizes $4+1$ and $5$, respectively). In both cases, one computes that the exponent of $N_U(C)/C$ is $3$, so there is no room for the cyclic group $D$ of order $9$.

There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.

As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.

Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.

On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)

From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$. Furthermore, $G$ acts regularly by order reasons.

The possible conjugacy classes of the $\text{GL}(4,3)$-parts of the generators of $C$ and $D$ are described by the Jordan normal form. Some computations and analysis of cases shows that there are no such groups you are looking for.

There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.

As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.

Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.

On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)

From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$.

View $G=C\rtimes D$ as a subgroup of $\text{GL}(5,3)$. From Jordan's normal form theorem, we see that $\text{GL}(5,3)$ contains two conjugacy classes of elements of order $9$. Let $U$ be the group of upper triangular matrices of $\text{GL}(5,3)$. Consider the two cases of $C$ (corresponding to the Jordan block sizes $3+1+1$ and $3+2$). In both cases, one computes that the exponent of $N_U(C)/C$ is $3$, so there is no room for the cyclic group $D$ of order $9$.