We go by induction, $n=1$ is trivial, suppose we are done for all smaller values of $n$. Now $n\ge 2$. $n\mid a^n-1$: let $d=ord_n(a)$. Then $d|n$ and $d|\varphi(n)$ (Euler-Fermat), hence $d<n$. $a^d\equiv 1\pmod n$. Suppose that $a^j+j\equiv a^k+k\pmod n$. Then let $j=dj_1+j_2$ and $k=dk_1+k_2$ where $0\le j_2,k_2<d$. We then get $a^{j_2}+j\equiv a^{k_2}+k\pmod n$ (1). Big idea: reduce this modulo $d$, then you get $a^{j_2}+j_2\equiv a^{k_2}+k_2\pmod d$. The induction hypothesis (we use the claim for $d$) implies that $j_2=k_2$. But then we get $a^{j_2}=a^{k_2}$, so (1) becomes $j\equiv k\pmod n$, and this implies $j=k$. So we are done for $n$ as well.

Is there any other solution?

This is problem B. 4720. from KöMaL.