First, let’s consider that we have a line segment on $\mathbb{R}$. By the definition of symmetry, we have $f:\mathbb{R}\to \mathbb{R}$ is symmetric if $d(f(a),f(b))=d(a,b)$ for all $a,b \in \mathbb{R}$.
Then the only symmetries possible in $\mathbb{R}$ would be the identity and the reflection about the center, which in our segments case would be $\frac {(b-a)} 2$.
Therefore the two possible linear transformations would be represented by the following matrices:
$P_0 =\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$ and $U_0 = \begin{bmatrix} -1&0\\0&-1 \end{bmatrix}$
Now consider the line segment on $\mathbb{R}^2$. The only symmetries possible would be rotation and reflection.
Generally on a plane, the dihedral group $D_n$ is generated via rotations $2k/n$ and reflections $2k/n$ where $k = 0,1,2,…,n-1$.
Since this is the dihedral group $D_2$ we can only have $0$ , , $0$ and .
Therefore, we have the symmetries that flip the plane around the line that contains the line segment, we get the symmetry that flips the plane around the perpendicular bisector and we get that planes complement and the identity matrix.
This leaves us with four transformations:
$P_0= \begin{bmatrix}1&0\\0&1\end{bmatrix}$ $P_\pi= \begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ $U_0 = \begin{bmatrix}-1&0\\0&1\end{bmatrix}$ $U_\pi= \begin{bmatrix}1&0\\0&-1\end{bmatrix}$
Therefore the symmetry group of a line segment in $\mathbb{R}^1$ has order $2$ and the symmetry group of a line segment in $\mathbb{R}^2$ has order $4$.