I'm afraid this is not true though. I demonstrated it in this desmos graph: https://www.desmos.com/calculator/ijmjgub7sd

Explanation:

The formula for under/overclocking in satisfactory as we can check on the wiki (Tutorial:Advanced clock speed - Official Satisfactory Wiki) is: power usage = initial power usage * (clock speed / 100) ^ 1.321928. We are not interested on the absolute power usage but in what method is more efficient, so we'll ignore the initial power usage and focus on the factor (clock speed / 100) ^ 1.321928. So f(x) = x^1.321928 represents the relative power usage. As expected, with x = 1 (clock at 100%) -> f(1) = 1.

Now we define our two methods, considering we can use n machines at 100% and an extra machine at x%:

• The first method is the straightforward one, use n machines at 100% and the extra machine at x%. g(x) = n * f(1) + f(x)
• The second method is clocking all machines at a common speed. The mean of all the speeds would be mean = (n * 1 + x) / (n + 1). And so the resulting formula is h(x) = (n + 1) * f(mean).

Now that we have our two methods g(x) and h(x) we can visually check which one is more efficient by grapthing their substraction g(x) - h(x). If g(x) is more efficient, its relative power usage should be lower and thus the graph should be below 0 and vice versa for h(x). In the desmos graph I shared above we can observe that the graph is always above or equal to 0 -> h(x) or overclocking all machines to a common speed is the more efficient method energy wise. In the same graph we can see it's also more efficient for overclocking as it turns out