Find all $n$ with $\!\bmod n\!:\, a\,$ invertible $\Rightarrow a^{-1} \equiv a,\,$ i.e. $a^2\equiv 1$

Find all positive integers $n$ such that $a \equiv a^{-1} \pmod{n}$ for all invertible $a$ modulo $n$.

I found that $n = 1,2,4,6,8,12,24$ satisfy this. How can we prove that these are all of them?

Equivalently, you are asking that $a^2\equiv 1\pmod{n}$ for any invertible $a$ mod $n$. By the Chinese remainder theorem, it suffices to consider the case that $n=p^d$ is a prime power. If $p$ is odd and $p^d\neq3$, then $a=2$ is invertible mod $p^d$ but $a^2=4\not\equiv 1\pmod{p^d}$. If $p=2$ and $d>3$, then $a=3$ is invertible mod $p^d$ but $a^2=9\not\equiv 1\pmod{p^d}$.

So if $n$ has the property you state, then the only possible odd prime factor of $n$ is $3$ (and $3^2\not\mid n$) and the highest power of $2$ dividing $n$ is at most $2^3$. That is, $n\mid 24$, which is exactly the numbers on your list together with $n=3$ (which is also an example and is missing from your list).

notice that this would imply $a^2\equiv 1$ for all $a$ coprime to $n$. Notice that $5$ does not work because $2^2$ is not $1\bmod 5$, we conclude that if $5|n$ then $n$ does not work.

Finally, if $5\nmid n$ then we should have that $5^2 \equiv 1 \bmod n$, which implies $n$ is a divisor of $24$.

The hypothesis implies that $U(n)$ has exponent $2$.

By the Chinese remainder theorem, the exponent of $U(n)$ is given by the Carmichael function $$ \lambda (n)=\operatorname {lcm}(\lambda (p_{1}^{{a_{1}}}),\;\lambda (p_{2}^{{a_{2}}}),\dots ,\lambda (p_{k}^{a_{k}})) $$ where $$ \lambda (p^a)= \begin{cases} \;\;\varphi(p^a) &\mbox{if } p\ne 2\\ \tfrac12\varphi(p^a)&\text{if }p=2 \end{cases} $$ Therefore, $\varphi(p_i^{a_i})=2$ for all odd prime factors of $n$, which implies that $p=3, a=1$ is the only possible candidate, and also that $n$ at most a factor $2^8$.

These restrictions give the list you have found.