A representation of $\mathbb{Z}$ is just a finite dimensional vector space over $k$ + an automorphism. The answer is probably quite complicated. For example, a representation of the one-dimensional Lie algebra $k$ is a finite dimensional vector space + an endomorphism. When $k$ is algebraically closed, the Tannakian fundamental group here is known, but it is complicated, namely, it is $Speck[t]×Speck[k]$. See question 21415 (answers of Milne/Ekedahl).

A representation of $\mathbb{Z}$ is just a finite dimensional vector space over $k$ + an automorphism. The answer is probably quite complicated. For example, a representation of the one-dimensional Lie algebra $k$ is a finite dimensional vector space + an endomorphism. When $k$ is algebraically closed, the Tannakian fundamental group here is known, but it is complicated, namely, it is $Speck[t]×Speck[k]$. See question 21415 (answers of Milne/Ekedahl).

The 1954 paper of Iwahori also considers automorphisms, so it may be possible to deduce the group from his results, as in the case of an endomorphism.

A representation of $\mathbb{Z}$ is just a finite dimensional vector space over $k$ + an endomorphism. This is the same as a representation of the one-dimensional Lie algebra $k$. When $k$ is algebraically closed, the Tannakian fundamental group is known, but it is complicated, namely, it is $Speck[t]×Speck[k]$. See question 21415 (answers of Milne/Ekedahl).

A representation of $\mathbb{Z}$ is just a finite dimensional vector space over $k$ + an automorphism. The answer is probably quite complicated. For example, a representation of the one-dimensional Lie algebra $k$ is a finite dimensional vector space + an endomorphism. When $k$ is algebraically closed, the Tannakian fundamental group here is known, but it is complicated, namely, it is $Speck[t]×Speck[k]$. See question 21415 (answers of Milne/Ekedahl).