The feasible region for the constraints $x-\mathrm{y} \geqslant 0, x-5 \mathrm{y} \leqslant-5, x \geqslant 0…

The constraints define the feasible region through four inequalities:

$x - y \ge 0 \implies y \le x$
$x - 5y \le -5 \implies y \ge \frac{1}{5}x + 1$
$x \ge 0$
$y \ge 0$

The boundary lines are $y = x$ and $y = \frac{1}{5}x + 1$, intersecting where $x = \frac{1}{5}x + 1$, yielding $\frac{4}{5}x = 1$, so $x = \frac{5}{4}$ and $y = \frac{5}{4}$.

Testing $(0,0)$ shows it violates $y \ge \frac{1}{5}x + 1$ since $0 \ge 1$ is false. The feasible region lies in the first quadrant, bounded below by $y = \frac{1}{5}x + 1$ and above by $y = x$, beginning at $(\frac{5}{4}, \frac{5}{4})$ and extending infinitely rightward.

Among the options, only Option C correctly depicts this unbounded region above $y = \frac{1}{5}x + 1$ and below $y = x$ in the first quadrant.

Final answer: $\boxed{C}$