It seems to me there is no solution. Proof:
The sum of the exterior angles of any polygon is $360^\circ$, so we need to find a set of numbers $\theta_0,\theta_1,\dots,\theta_k=\theta_0$ such that $|\theta_i-\theta_{i-1}|=5^\circ$ and $\sum_1^k\theta_i=360^\circ$. We also know that $\min\{180^\circ-\theta_i\}=120^\circ$, so $\theta_i\leq 60^\circ$ and $(\exists i)[\theta_i=60^\circ]$. WLOG, we can let $\theta_0=60^\circ$. Noting that $\theta_i/5^\circ=n_i\in\mathbb Z$, the constraint $|n_i-n_{i-1}|=1$ means that if $n_i$ is even, $n_{i+1}$ is odd and vice-versa so that $n_i$ is even iff $i$ is even since $n_0=12$ by induction.
Thus, if $k$ is odd, then $n_k=n_0$ is odd and even, a contradiction. Thus $k$ is even. Additionally, since $n_0+n_1,n_2+n_3,\dots$ are all odd, $\sum_1^k n_i$ will be odd if $k/2$ is odd, which is a contradiction since $\sum_1^k n_i=72$. Thus $k$ is a multiple of $4$.
Now $n_i\leq 12\Rightarrow\sum_1^k n_i=72\leq12k\Rightarrow k\geq 6,$ and $|n_i-n_{i-1}|=1\Rightarrow n_i\geq12-\min(i,n-i)\Rightarrow$
$72=\sum_1^k n_i=n_0+n_{k/2}+\sum_1^{k/2-1} (n_i+n_{n-i})\geq 24-k/2+2\sum_1^{k/2-1} (12-i)=12k-(k/2)^2,$
so $k\leq12(2-\sqrt2)\approx7.02$. There are no multiples of 4 in the range $k\in[6,7.02]$, so this has no solution.