By intersecting a line $r$ with a line $s$ passing through $(x,y)$ and perpendicular to $r$, we have:
$$
(x_0,y_0) + (a,b)\,u = (x,y) + (-b,a)\,v
\quad \quad \Leftrightarrow \quad \quad
\begin{cases}
u = \frac{a(x-x_0)+b(y-y_0)}{a^2+b^2} \\
v = \frac{b(x-x_0)-a(y-y_0)}{a^2+b^2} \\
\end{cases}\,.
$$
Therefore, the point $(x',y')$ symmetrical of $(x,y)$ with respect to $r$ is:
$$
\boxed{(x',y') = (x,y) + (-b,a)\left(2\frac{b(x-x_0)-a(y-y_0)}{a^2+b^2}\right)}\,.
$$
In your case, since $\color{magenta}{r}$ passes through $(x_0,y_0)=(0,0)$ and has director vector $(a,b)=(5,4)$:
$$
(x',y') = \left(\frac{9x+40y}{41},\frac{40x-9y}{41}\right)
$$
from which it follows that the points of the curve $\color{blue}{(x,y) = (t,e^{t/2}/5)}$ will be symmetrized in:
$$
\color{red}{(x,y) = \left(\frac{9t+8e^{t/2}}{41},\frac{200t-9e^{t/2}}{205}\right)}
$$
where as $t$ varies in $\mathbb{R}$ we obtain the points of the blue and red curves of our interest.
$\quad\quad\quad\quad\quad$
On the other hand, if for some reason a Cartesian equation is preferred:
$$
\color{blue}{y = \frac{1}{5}e^{\frac{x}{2}}}
\quad \quad \Rightarrow \quad \quad
\color{red}{\frac{40x-9y}{41} = \frac{1}{5}e^{\frac{9x+40y}{82}}}
$$
where to make $\color{red}{y}$ explicit it's necessary to use the Lambert W function:
$$
\color{red}{y = \frac{40}{9}x - \frac{41}{20}\,\mathcal{W}\left(\frac{4}{9}e^{\frac{41}{18}x}\right)}.
$$
This proves that it isn't always possible to get the symmetric function in an elementary way.
