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.