The affine group scheme $G$ you describe is not finite type. It is possible to describe $G$ explicitly.

For $R$ a $k$-algebra, the $R$-points of $G$ are the tensor functorial automorphisms of $\mathrm{Rep}_k(\mathbb{Z})\to R\mathrm{-mod}$, $V\mapsto V\otimes R$. Let $\varphi$ be such an automorphism.

For $\lambda\in k^\times$, let $V_\lambda$ be the $1$-dimensiona $\mathbb{Z}$-representation where the generator acts by multiplication by $\lambda$. Then $\varphi_{V_\lambda}$ is multiplication by some constant $c_\lambda\in R^\times$, and the isomorphism $V_{\lambda}\otimes V_{\mu}\cong V_{\lambda\mu}$ implies $\lambda\mapsto c_\lambda$ must be a group homomorphism $k^\times\to R^\times$.

For $n$ a positive integer, let $W_n$ be the $n$-dimensional $\mathbb{Z}$-representation where the generator acts by an $n\times n$ Jordan block with 1's on the diagonal. You can check that $\varphi_{W_n}$ acts by a matrix of the form $$ \left[\begin{array}{ccccc}1&b_1&b_2&\ldots&b_{n-1}\\ 0&1&b_1&\ldots&b_{n-2}\\ \vdots&&\ddots&&\\ 0&0&0&\ldots&1 \end{array}\right]. $$ The elements $b_i\in R$ do not depend on $n$, and you can check that they satisfy relations that say $b_i$ acts like ${x\choose i}$ for $x$ an indeterminate: $$ \tag{$\star$} b_i b_j={i+j\choose i} b_{i+j}. $$

Conversely, given a group homomorphism $k^\times\to R^\times$, $\lambda\mapsto c_\lambda$ and elements $b_i\in R$ satisfying $(\star)$, we can define an automorphism of $V\otimes R$ for each $\mathbb{Z}$-rep $V$ in the following way. We can think of a $\mathbb{Z}$-rep as a square matrix $M$ (giving the action of the generator with respect to some choice of basis). As a function of $n$, the entries of $M^n$ can all be written as $k$-linear combinations of terms of the form $\lambda^n {n\choose j}$, for $\lambda\in k^\times$ and $j\in\mathbb{Z}_{\geq 0}$. We replace $\lambda^n$ by $c_\lambda$ and ${n\choose j}$ by $b_j$ to get a square matrix over $R$, and this is our $R$-automorphism of the forgetful functor.

We have shown that $$ G(R)=\{(\varphi,(b_i)):\varphi:k^\times\to R^\times,\,\,b_i\in R\text{ satisfying ($\star$)}\}. $$ There is a group homomorphism $\mathbb{Z}\to G(k)$, sending $n\in \mathbb{Z}$ to $(\varphi,(b_i))$ with $\varphi(\lambda)=\lambda^n$ and $b_i={n\choose i}$. The coordinate ring of $G$ is $$ k[k^\times]\otimes k\big[{x\choose i}\big], $$ where $k[k^\times]$ is the group algebra of $k^\times$, and $k\big[{x\choose i}\big]$ is the algebra generated by formal symbols ${x\choose i}$, subject to the relations $$ {x\choose i}{x\choose j}={i+j\choose i}{x\choose i+j}. $$ In characteristic $0$, $k\big[{x\choose i}\big]=k[x]$.

The affine group scheme $G$ you describe is not finite type. It is possible to describe $G$ explicitly.

For $R$ a $k$-algebra, the $R$-points of $G$ are the tensor functorial automorphisms of $\mathrm{Rep}_k(\mathbb{Z})\to R\mathrm{-mod}$, $V\mapsto V\otimes R$. Let $\varphi$ be such an automorphism.

For $\lambda\in k^\times$, let $V_\lambda$ be the $1$-dimensiona $\mathbb{Z}$-representation where the generator acts by multiplication by $\lambda$. Then $\varphi_{V_\lambda}$ is multiplication by some constant $c_\lambda\in R^\times$, and the isomorphism $V_{\lambda}\otimes V_{\mu}\cong V_{\lambda\mu}$ implies $\lambda\mapsto c_\lambda$ must be a group homomorphism $k^\times\to R^\times$.

For $n$ a positive integer, let $W_n$ be the $n$-dimensional $\mathbb{Z}$-representation where the generator acts by an $n\times n$ Jordan block with 1's on the diagonal. You can check that $\varphi_{W_n}$ acts by a matrix of the form $$ \left[\begin{array}{ccccc}1&b_1&b_2&\ldots&b_{n-1}\\ 0&1&b_1&\ldots&b_{n-2}\\ \vdots&&\ddots&&\\ 0&0&0&\ldots&1 \end{array}\right]. $$ The elements $b_i\in R$ do not depend on $n$, and you can check that they satisfy the relations satisfied by ${x\choose i}$ for $x$ an indeterminate: for each $i$, $j$, there are integers $c_{i,j,k}$ such that ${x\choose i}{x\choose j}=\sum_k c_{i,j,k}{x\choose k}$ for all $x$, and we will have $$\tag{$\star$}b_ib_j=\sum_k c_{i,j,k}b_k.$$

Conversely, given a group homomorphism $k^\times\to R^\times$, $\lambda\mapsto c_\lambda$ and elements $b_i\in R$ satisfying the appropriate relations, we can define an automorphism of $V\otimes R$ for each $\mathbb{Z}$-rep $V$ in the following way. We can think of a $\mathbb{Z}$-rep as a square matrix $M$ (giving the action of the generator with respect to some choice of basis). As a function of $n$, the entries of $M^n$ can all be written as $k$-linear combinations of terms of the form $\lambda^n {n\choose j}$, for $\lambda\in k^\times$ and $j\in\mathbb{Z}_{\geq 0}$. We replace $\lambda^n$ by $c_\lambda$ and ${n\choose j}$ by $b_j$ to get a square matrix over $R$, and this is our $R$-automorphism of the forgetful functor.

We have shown that $$ G(R)=\{(\varphi,(b_i)):\varphi:k^\times\to R^\times,\,\,b_i\in R\text{ satisfying ($\star$)}\}. $$ There is a group homomorphism $\mathbb{Z}\to G(k)$, sending $n\in \mathbb{Z}$ to $(\varphi,(b_i))$ with $\varphi(\lambda)=\lambda^n$ and $b_i={n\choose i}$. The coordinate ring of $G$ is $$ k[k^\times]\otimes k\big[{x\choose i}\big], $$ where $k[k^\times]$ is the group algebra of $k^\times$, and $k\big[{x\choose i}\big]$ is the algebra generated by formal symbols ${x\choose i}$, subject to the relations $(\star)$. In characteristic $0$, $k\big[{x\choose i}\big]=k[x]$.

The affine group scheme $G$ you describe is not finite type. It is possible to describe the coordinate ring of $G$ explicitly.

For $R$ a $k$-algebra, the $R$-points of $G$ are the tensor functorial automorphisms of $\mathrm{Rep}_k(\mathbb{Z})\to R\mathrm{-mod}$, $V\mapsto V\otimes R$. Let $\varphi$ be such an automorphism.

For $\lambda\in k^\times$, let $V_\lambda$ be the $1$-dimensiona $\mathbb{Z}$-representation where the generator acts by multiplication by $\lambda$. Then $\varphi_{V_\lambda}$ is multiplication by some constant $c_\lambda\in R^\times$, and the isomorphism $V_{\lambda}\otimes V_{\mu}\cong V_{\lambda\mu}$ implies $\lambda\mapsto c_\lambda$ must be a group homomorphism $k^\times\to R^\times$.

For $n$ a positive integer, let $W_n$ be the $n$-dimensional $\mathbb{Z}$-representation where the generator acts by an $n\times n$ Jordan block with 1's on the diagonal. You can check that $\varphi_{W_n}$ acts by a matrix of the form $$ \left[\begin{array}{ccccc}1&b_1&b_2&\ldots&b_{n-1}\\ 0&1&b_1&\ldots&b_{n-2}\\ \vdots&&\ddots&&\\ 0&0&0&\ldots&1 \end{array}\right]. $$ The constants $b_i\in R$ do not depend on $n$, and you can check that they satisfy relations that say $b_i$ acts like ${x\choose i}$ for $x$ an indeterminate: $$ \tag{$\star$} b_i b_j={i+j\choose i} b_{i+j}. $$

Conversely, given a group homomorphism $k^\times\to R^\times$, $\lambda\mapsto c_\lambda$ and elements $b_i\in R$ satisfying $(\star)$, we can define an automorphism of $V\otimes R$ for each $\mathbb{Z}$-rep $V$ in the following way. We can think of a $\mathbb{Z}$-rep as a square matrix $M$ (giving the action of the generator with respect to some choice of basis). As a function of $n$, the entries of $M^n$ can all be written as $k$-linear combinations of terms of the form $\lambda^n {n\choose j}$, for $\lambda\in k^\times$ and $j\in\mathbb{Z}_{\geq 0}$. We replace $\lambda^n$ by $c_\lambda$ and ${n\choose j}$ by $b_j$ to get a square matrix over $R$, and this is our $R$-automorphism of the forgetful functor.

We have described the $R$-points of $G$, and we can read off that the coordinate ring is $$ k[k^\times]\otimes k\big[{x\choose i}\big], $$ where $k[k^\times]$ is the group algebra of $k^\times$, and $k\big[{x\choose i}\big]$ is the algebra generated by formal symbols ${x\choose i}$, subject to the relations $$ {x\choose i}{x\choose j}={i+j\choose i}{x\choose i+j}. $$ In characteristic $0$, $k\big[{x\choose i}\big]=k[x]$.

The affine group scheme $G$ you describe is not finite type. It is possible to describe $G$ explicitly.

For $R$ a $k$-algebra, the $R$-points of $G$ are the tensor functorial automorphisms of $\mathrm{Rep}_k(\mathbb{Z})\to R\mathrm{-mod}$, $V\mapsto V\otimes R$. Let $\varphi$ be such an automorphism.

For $\lambda\in k^\times$, let $V_\lambda$ be the $1$-dimensiona $\mathbb{Z}$-representation where the generator acts by multiplication by $\lambda$. Then $\varphi_{V_\lambda}$ is multiplication by some constant $c_\lambda\in R^\times$, and the isomorphism $V_{\lambda}\otimes V_{\mu}\cong V_{\lambda\mu}$ implies $\lambda\mapsto c_\lambda$ must be a group homomorphism $k^\times\to R^\times$.

For $n$ a positive integer, let $W_n$ be the $n$-dimensional $\mathbb{Z}$-representation where the generator acts by an $n\times n$ Jordan block with 1's on the diagonal. You can check that $\varphi_{W_n}$ acts by a matrix of the form $$ \left[\begin{array}{ccccc}1&b_1&b_2&\ldots&b_{n-1}\\ 0&1&b_1&\ldots&b_{n-2}\\ \vdots&&\ddots&&\\ 0&0&0&\ldots&1 \end{array}\right]. $$ The elements $b_i\in R$ do not depend on $n$, and you can check that they satisfy relations that say $b_i$ acts like ${x\choose i}$ for $x$ an indeterminate: $$ \tag{$\star$} b_i b_j={i+j\choose i} b_{i+j}. $$

Conversely, given a group homomorphism $k^\times\to R^\times$, $\lambda\mapsto c_\lambda$ and elements $b_i\in R$ satisfying $(\star)$, we can define an automorphism of $V\otimes R$ for each $\mathbb{Z}$-rep $V$ in the following way. We can think of a $\mathbb{Z}$-rep as a square matrix $M$ (giving the action of the generator with respect to some choice of basis). As a function of $n$, the entries of $M^n$ can all be written as $k$-linear combinations of terms of the form $\lambda^n {n\choose j}$, for $\lambda\in k^\times$ and $j\in\mathbb{Z}_{\geq 0}$. We replace $\lambda^n$ by $c_\lambda$ and ${n\choose j}$ by $b_j$ to get a square matrix over $R$, and this is our $R$-automorphism of the forgetful functor.

We have shown that $$ G(R)=\{(\varphi,(b_i)):\varphi:k^\times\to R^\times,\,\,b_i\in R\text{ satisfying ($\star$)}\}. $$ There is a group homomorphism $\mathbb{Z}\to G(k)$, sending $n\in \mathbb{Z}$ to $(\varphi,(b_i))$ with $\varphi(\lambda)=\lambda^n$ and $b_i={n\choose i}$. The coordinate ring of $G$ is $$ k[k^\times]\otimes k\big[{x\choose i}\big], $$ where $k[k^\times]$ is the group algebra of $k^\times$, and $k\big[{x\choose i}\big]$ is the algebra generated by formal symbols ${x\choose i}$, subject to the relations $$ {x\choose i}{x\choose j}={i+j\choose i}{x\choose i+j}. $$ In characteristic $0$, $k\big[{x\choose i}\big]=k[x]$.