Mathematically, if writing the linear model as $y = X\beta + \epsilon$, where $X = \begin{bmatrix}e & x_1 & x_2 & \cdots & x_p\end{bmatrix}$, $\beta = \begin{bmatrix}\beta_0 & \beta_1 & \beta_2 & \cdots & \beta_p\end{bmatrix}'$. The standard error of $\hat{\beta}_j, 1 \leq j \leq p$, denoted by $\hat{\sigma}_{\hat{\beta}_j}$, is then given by \begin{align} \hat{\sigma}\sqrt{e_j'(X'X)^{-1}e_j}, \tag{1} \end{align} where $e_j$ is a $(p+1)$-long column vector whose $(j + 1)$-st entry is $1$ and all the other entries $0$, $\hat{\sigma} = (n - p - 1)^{-1}y'(I - H)y$, $H = X(X'X)^{-1}X'$. In other words, $\hat{\sigma}_{\hat{\beta}_j}$ is the square root of the $(j + 1, j + 1)$ diagonal entry of the matrix $\hat{\sigma}^2(X'X)^{-1}$. Therefore, if all the diagonal entries of $(X'X)^{-1}$ are the same (or block-wise same), then you would see identical (or block-wise identical) standard errors of OLS estimates.

Although you did not include details of the input data in your post, the output implies that your data probably came from a designed experiment, where the columns of $X$ are mutually orthogonal. In this case, the diagonal entries of $(X'X)^{-1}$ could be identical. For example, suppose \begin{align} X = \begin{bmatrix} 1_9 & 1_9 & 0 & 0 \\ 1_9 & 0 & 1_9 & 0 \\ 1_9 & 0 & 0 & 1_9 \\ 1_9 & 0 & 0 & 0 \end{bmatrix}, \end{align} where $1_9$ is a $9$-long column vector consisting of all ones. It is then easy to verify that \begin{align} X'X = \begin{bmatrix} 36 & 9 & 9 & 9 \\ 9 & 9 & 0 & 0 \\ 9 & 0 & 9 & 0 \\ 9 & 0 & 0 & 9 \end{bmatrix}, \quad (X'X)^{-1} = \frac{1}{9}\begin{bmatrix} 1 & -1 & -1 & -1 \\ -1 & 2 & 1 & 1 \\ -1 & 1 & 2 & 1 \\ -1 & 1 & 1 & 2 \end{bmatrix}. \end{align} It can be seen that all the $(2, 2), (3, 3), (4, 4)$ diagonal entries equal to $2$, yielding the same standard errors of $\hat{\beta}_1, \hat{\beta}_2$ and $\hat{\beta}_3$.

To analyze your particular case, you can use model.matrix command to print out your design matrix $X$ and compute $(X'X)^{-1}$ to verify the reason.

Mathematically, if writing the linear model as $y = X\beta + \epsilon$, where $X = \begin{bmatrix}e & x_1 & x_2 & \cdots & x_p\end{bmatrix}$, $\beta = \begin{bmatrix}\beta_0 & \beta_1 & \beta_2 & \cdots & \beta_p\end{bmatrix}'$. The standard error of $\hat{\beta}_j, 1 \leq j \leq p$, denoted by $\hat{\sigma}_{\hat{\beta}_j}$, is then given by \begin{align} \hat{\sigma}\sqrt{e_j'(X'X)^{-1}e_j}, \tag{1} \end{align} where $e_j$ is a $(p+1)$-long column vector whose $(j + 1)$-st entry is $1$ and all the other entries $0$, $\hat{\sigma} = (n - p - 1)^{-1}y'(I - H)y$, $H = X(X'X)^{-1}X'$. In other words, $\hat{\sigma}_{\hat{\beta}_j}$ is the square root of the $(j + 1, j + 1)$ diagonal entry of the matrix $\hat{\sigma}^2(X'X)^{-1}$. Therefore, if all the diagonal entries of $(X'X)^{-1}$ are the same (or block-wise same), then you would see identical (or block-wise identical) standard errors of OLS estimates.

Although you did not include details of the input data in your post, the output implies that your data probably came from a designed experiment, where the columns of $X$ are mutually orthogonal. In this case, the diagonal entries of $(X'X)^{-1}$ could be identical. For example, suppose \begin{align} X = \begin{bmatrix} 1_9 & 1_9 & 0 & 0 \\ 1_9 & 0 & 1_9 & 0 \\ 1_9 & 0 & 0 & 1_9 \\ 1_9 & 0 & 0 & 0 \end{bmatrix}, \end{align} where $1_9$ is a $9$-long column vector consisting of all ones. It is then easy to verify that \begin{align} X'X = \begin{bmatrix} 36 & 9 & 9 & 9 \\ 9 & 9 & 0 & 0 \\ 9 & 0 & 9 & 0 \\ 9 & 0 & 0 & 9 \end{bmatrix}, \quad (X'X)^{-1} = \frac{1}{9}\begin{bmatrix} 1 & -1 & -1 & -1 \\ -1 & 2 & 1 & 1 \\ -1 & 1 & 2 & 1 \\ -1 & 1 & 1 & 2 \end{bmatrix}. \end{align} It can be seen that all the $(2, 2), (3, 3), (4, 4)$ diagonal entries of $(X'X)^{-1}$ equal to $\frac{2}{9}$, yielding the same standard errors of $\hat{\beta}_1, \hat{\beta}_2$ and $\hat{\beta}_3$.

To analyze your particular case, you can use model.matrix command to print out your design matrix $X$ and compute $(X'X)^{-1}$ to verify the reason.

Mathematically, if writing the linear model as $y = X\beta + \epsilon$, where $X = \begin{bmatrix}e & x_1 & x_2 & \cdots & x_p\end{bmatrix}$, $\beta = \begin{bmatrix}\beta_0 & \beta_1 & \beta_2 & \cdots & \beta_p\end{bmatrix}$. The standard error of $\hat{\beta}_j, 1 \leq j \leq p$, denoted by $\hat{\sigma}_{\hat{\beta}_j}$, is then given by \begin{align} \hat{\sigma}\sqrt{e_j'(X'X)^{-1}e_j}, \tag{1} \end{align} where $e_j$ is a $(p+1)$-long column vector whose $(j + 1)$-st entry is $1$ and all the other entries $0$, $\hat{\sigma} = (n - p - 1)^{-1}y'(I - H)y$, $H = X(X'X)^{-1}X'$. In other words, $\hat{\sigma}_{\hat{\beta}_j}$ is the square root of the $(j + 1, j + 1)$ diagonal entry of the matrix $\hat{\sigma}^2(X'X)^{-1}$. Therefore, if all the diagonal entries of $(X'X)^{-1}$ are the same (or block-wise same), then you would see identical (or block-wise identical) standard errors of OLS estimates.

Although you did not include details of the input data in your post, the output implies that your data probably came from a designed experiment, where the columns of $X$ are mutually orthogonal. In this case, the diagonal entries of $(X'X)^{-1}$ could be identical. For example, suppose \begin{align} X = \begin{bmatrix} e_9 & e_9 & 0 & 0 \\ e_9 & 0 & e_9 & 0 \\ e_9 & 0 & 0 & e_9 \\ e_9 & 0 & 0 & 0 \end{bmatrix}, \end{align} where $e_9$ is a $9$-long column vector consisting of all ones. It is then easy to verify that \begin{align} X'X = \begin{bmatrix} 36 & 9 & 9 & 9 \\ 9 & 9 & 0 & 0 \\ 9 & 0 & 9 & 0 \\ 9 & 0 & 0 & 9 \end{bmatrix}, \quad (X'X)^{-1} = \frac{1}{9}\begin{bmatrix} 1 & -1 & -1 & -1 \\ -1 & 2 & 1 & 1 \\ -1 & 1 & 2 & 1 \\ -1 & 1 & 1 & 2 \end{bmatrix}. \end{align} It can be seen that all the $(2, 2), (3, 3), (4, 4)$ diagonal entries equal to $2$, yielding the same standard errors of $\hat{\beta}_1, \hat{\beta}_2$ and $\hat{\beta}_3$.

To analyze your particular case, you can use model.matrix command to print out your design matrix $X$ and compute $(X'X)^{-1}$ to verify the reason.

Mathematically, if writing the linear model as $y = X\beta + \epsilon$, where $X = \begin{bmatrix}e & x_1 & x_2 & \cdots & x_p\end{bmatrix}$, $\beta = \begin{bmatrix}\beta_0 & \beta_1 & \beta_2 & \cdots & \beta_p\end{bmatrix}'$. The standard error of $\hat{\beta}_j, 1 \leq j \leq p$, denoted by $\hat{\sigma}_{\hat{\beta}_j}$, is then given by \begin{align} \hat{\sigma}\sqrt{e_j'(X'X)^{-1}e_j}, \tag{1} \end{align} where $e_j$ is a $(p+1)$-long column vector whose $(j + 1)$-st entry is $1$ and all the other entries $0$, $\hat{\sigma} = (n - p - 1)^{-1}y'(I - H)y$, $H = X(X'X)^{-1}X'$. In other words, $\hat{\sigma}_{\hat{\beta}_j}$ is the square root of the $(j + 1, j + 1)$ diagonal entry of the matrix $\hat{\sigma}^2(X'X)^{-1}$. Therefore, if all the diagonal entries of $(X'X)^{-1}$ are the same (or block-wise same), then you would see identical (or block-wise identical) standard errors of OLS estimates.

Although you did not include details of the input data in your post, the output implies that your data probably came from a designed experiment, where the columns of $X$ are mutually orthogonal. In this case, the diagonal entries of $(X'X)^{-1}$ could be identical. For example, suppose \begin{align} X = \begin{bmatrix} 1_9 & 1_9 & 0 & 0 \\ 1_9 & 0 & 1_9 & 0 \\ 1_9 & 0 & 0 & 1_9 \\ 1_9 & 0 & 0 & 0 \end{bmatrix}, \end{align} where $1_9$ is a $9$-long column vector consisting of all ones. It is then easy to verify that \begin{align} X'X = \begin{bmatrix} 36 & 9 & 9 & 9 \\ 9 & 9 & 0 & 0 \\ 9 & 0 & 9 & 0 \\ 9 & 0 & 0 & 9 \end{bmatrix}, \quad (X'X)^{-1} = \frac{1}{9}\begin{bmatrix} 1 & -1 & -1 & -1 \\ -1 & 2 & 1 & 1 \\ -1 & 1 & 2 & 1 \\ -1 & 1 & 1 & 2 \end{bmatrix}. \end{align} It can be seen that all the $(2, 2), (3, 3), (4, 4)$ diagonal entries equal to $2$, yielding the same standard errors of $\hat{\beta}_1, \hat{\beta}_2$ and $\hat{\beta}_3$.

To analyze your particular case, you can use model.matrix command to print out your design matrix $X$ and compute $(X'X)^{-1}$ to verify the reason.