Assuming that each point must be visited only once in the circuit (otherwise it would be trivial) the answer is no. Moving along the diagonals implies that each movement is from a vertex to a midpoint, or vice versa. So the circuit must be an alternation of vertices and midpoints. The vertices are <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>8</mn></mrow><annotation encoding="application/x-tex">8</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">8</span></span></span></span>, the midpoints are <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>6</mn></mrow><annotation encoding="application/x-tex">6</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">6</span></span></span></span>. So, even starting from a vertex, after visiting the last midpoint, two vertices remain and there can be no way to visit both of them.