There is no example with eigenvalues $-1$. More generally, suppose that $\mathfrak{g}$ is a Lie algebra, and $\alpha$ is an automorphism of order $2$ whose fixed subspace is trivial. Then I claim that $\mathfrak{g}$ is abelian.

Proof: Since $\alpha^2=\mathrm{Id}$ and $\alpha$ has no fixed points, we must have $\alpha = -\mathrm{Id}$. For any $u$ and $v$ in $\mathfrak{g}$, we have $[\alpha u, \alpha v] = \alpha [u, v]$, since $\alpha$ is an automorphism. But we just showed $\alpha = - \mathrm{Id}$, so this shows $[-u, -v] = - [u,v]$ and thus $[u,v]=0$. Since $u$ and $v$ were arbitrary, this shows that $\mathfrak{g}$ is abelian. QED

However, your bulleted conditions do not force $\alpha$ to have all eigenvalues $-1$. I will first construct an example over $\mathbb{C}$, then modify it to work over $\mathbb{R}$.

Let $\mathfrak{n}$ be the Lie algebra of strictly upper triangular $3 \times 3$ complex matrices. Let $\omega$ be a primitive cube root of unity and consider the automorphism
$$\alpha : \begin{pmatrix} 0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix} \mapsto
\begin{pmatrix} 0 & \omega x & \omega^2 y \\ 0 & 0 & \omega z \\ 0 & 0 & 0 \end{pmatrix}$$

This has order $3$, and has no nonzero fixed points. If you like, you can think of this as conjugation by the diagonal matrix $\mathrm{diag}(1, \omega, \omega^2)$.

To make an example over $\mathbb{R}$, just look at the same Lie algebra with the same automorphism and forget that it is a complex vector space, remembering only the real vector space structure. So it is now a $6$-dimensional real Lie algebra, with an automorphism of order $3$, and no fixed points.

Finally, I need to give an $\alpha$-stable lattice in $\mathfrak{n}$. Consider the lattice of matrices as above where each of $x$, $y$ and $z$ are in the ring $\mathbb{Z}[\omega]$ (sometimes called the Eisenstein integers).

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