5. The solution to the system of equations $\left\{\begin{array}{l}x=\mathrm{e} \ln y, \\ y=\mathrm{e} \ln z, \\ z=\mathrm{e} \ln x\end{array}\right.$ is
5. $x=y=z=\mathrm{e}$.

The original system of equations can be transformed into
$$
y=\mathrm{e}^{\frac{x}{0}}, z=\mathrm{e}^{\frac{y}{0}}, x=\mathrm{e}^{\frac{x}{\frac{1}{x}}} \text {. }
$$

Assume $x>y$.
From the first two equations, we get $y>z$; from the last two equations, we get $z>x$. Hence, $y>x$, which contradicts the assumption. Therefore, $x \leqslant y$.
Similarly, $x \geqslant y$. Thus, $x=y$.
Similarly, $y=z$. Therefore, $x=y=z$.
Hence, $x=\mathrm{e}^{\frac{x}{\circ}} \Rightarrow x=\mathrm{e} \Rightarrow x=y=z=\mathrm{e}$.