It's easier to describe the endomorphism ring first, then take its group of units. In general (in any additive category!) the endomorphism ring of $A \oplus B$ is the ring of "matrices"

$$\begin{bmatrix} \operatorname{End}(A) & \operatorname{Hom}(B, A) \\ \operatorname{Hom}(A, B) & \operatorname{End}(B) \end{bmatrix}.$$

Here we get lucky: $\operatorname{Hom}(\mathbb{T}, \mathbb{R}) = 0$ so one of the off-diagonal terms disappears. Using the fact that $\operatorname{End}(\mathbb{T}) \cong \mathbb{Z}$ and $\operatorname{Hom}(\mathbb{R}, \mathbb{T}) \cong \mathbb{R}$, we get that $\operatorname{End}(\mathbb{R}^n \oplus \mathbb{T}^m)$ is

$$\begin{bmatrix} M_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & M_m(\mathbb{Z}) \end{bmatrix}.$$

Because our "matrices" are now "lower triangular" the group of units is just given by taking the group of units along the diagonal; we get that the automorphism group is

$$\begin{bmatrix} \operatorname{GL}_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & \operatorname{GL}_m(\mathbb{Z}) \end{bmatrix}.$$

(Below we work with continuous endomorphisms / automorphisms, i.e. this is all taking place in the category of locally compact Hausdorff abelian groups and continuous homomorphisms.)

It's easier to describe the endomorphism ring first, then take its group of units. In general (in any additive category!) the endomorphism ring of $A \oplus B$ is the ring of "matrices"

$$\begin{bmatrix} \operatorname{End}(A) & \operatorname{Hom}(B, A) \\ \operatorname{Hom}(A, B) & \operatorname{End}(B) \end{bmatrix}.$$

Here we get lucky: $\operatorname{Hom}(\mathbb{T}, \mathbb{R}) = 0$ so one of the off-diagonal terms disappears. Using the fact that $\operatorname{End}(\mathbb{T}) \cong \mathbb{Z}$ and $\operatorname{Hom}(\mathbb{R}, \mathbb{T}) \cong \mathbb{R}$, we get that $\operatorname{End}(\mathbb{R}^n \oplus \mathbb{T}^m)$ is

$$\begin{bmatrix} M_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & M_m(\mathbb{Z}) \end{bmatrix}.$$

Because our "matrices" are now "lower triangular" the group of units is just given by taking the group of units along the diagonal; we get that the automorphism group is

$$\begin{bmatrix} \operatorname{GL}_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & \operatorname{GL}_m(\mathbb{Z}) \end{bmatrix}.$$

It's easier to describe the endomorphism ring first, then take its group of units. In general (in any additive category!) the endomorphism ring of $A \oplus B$ is the ring of "matrices"

$$\begin{bmatrix} \text{End}(A) & \text{Hom}(B, A) \\ \text{Hom}(A, B) & \text{End}(B). \end{bmatrix}.$$

Here we get lucky: $\text{Hom}(\mathbb{T}, \mathbb{R}) = 0$ so one of the off-diagonal terms disappears. Using the fact that $\text{End}(\mathbb{T}) \cong \mathbb{Z}$ and $\text{Hom}(\mathbb{R}, \mathbb{T}) \cong \mathbb{R}$, we get that $\text{End}(\mathbb{R}^n \oplus \mathbb{T}^m)$ is

$$\begin{bmatrix} M_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & M_m(\mathbb{Z}) \end{bmatrix}.$$

Because our "matrices" are now "lower triangular" the group of units is just given by taking the group of units along the diagonal; we get that the automorphism group is

$$\begin{bmatrix} GL_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & GL_m(\mathbb{Z}) \end{bmatrix}.$$

It's easier to describe the endomorphism ring first, then take its group of units. In general (in any additive category!) the endomorphism ring of $A \oplus B$ is the ring of "matrices"

$$\begin{bmatrix} \operatorname{End}(A) & \operatorname{Hom}(B, A) \\ \operatorname{Hom}(A, B) & \operatorname{End}(B) \end{bmatrix}.$$

Here we get lucky: $\operatorname{Hom}(\mathbb{T}, \mathbb{R}) = 0$ so one of the off-diagonal terms disappears. Using the fact that $\operatorname{End}(\mathbb{T}) \cong \mathbb{Z}$ and $\operatorname{Hom}(\mathbb{R}, \mathbb{T}) \cong \mathbb{R}$, we get that $\operatorname{End}(\mathbb{R}^n \oplus \mathbb{T}^m)$ is

$$\begin{bmatrix} M_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & M_m(\mathbb{Z}) \end{bmatrix}.$$

Because our "matrices" are now "lower triangular" the group of units is just given by taking the group of units along the diagonal; we get that the automorphism group is

$$\begin{bmatrix} \operatorname{GL}_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & \operatorname{GL}_m(\mathbb{Z}) \end{bmatrix}.$$