I'm trying to find the geodesic that connects the identity with some circulant, symmetric matrix $U\in\mathrm{GL}(2N,\mathbb{R})$, meaning we have
\begin{align}
	U=\left(\begin{array}{ccc}
	u_1 & u_2 & \cdots\\
	u_2 & u_1 & \cdots\\
	\vdots & \vdots & \ddots
	\end{array}\right)\quad\text{with}\quad b_{N-i}=b_{1+i}\,.
\end{align}
Given two matrices $A$ and $B$ in the Lie algebra $\mathrm{gl}(2N,\mathbb{R})$, their inner product is defined by
\begin{align}
g(A,B)=\mathrm{tr}\left[\left(\begin{array}{ccccc}
	a_{11} & \alpha\,a_{12} & \cdots & \alpha^2\,a_{1(N-1)}& \alpha\,a_{1N}\\
	\alpha\,a_{21} & a_{22} & \cdots & \alpha^3\,a_{2(N-1)}& \alpha^2\,a_{1N}\\
	\vdots & \vdots & \ddots & \vdots & \vdots
	\end{array}\right)\left(\begin{array}{ccc}
	b_{11} & \alpha\,b_{12} & \cdots & \alpha^2\,b_{1(N-1)}& \alpha\,b_{1N}\\
	\alpha\,b_{21} & b_{22} & \cdots & \alpha^3\,b_{2(N-1)}& \alpha^2\,b_{1N}\\
	\vdots & \vdots & \ddots & \vdots & \vdots
	\end{array}\right)\right]\,.
\end{align}
For other tangent vectors, we require that the metric is right-invariant.

The metric has the same invariance as circulant & symmetric matrices. I want to argue that the geodesic itself $U(t)$ is itself circulant & symmetric for every $t$. Is there a simple way to prove this?