Given that the convolution $f*g$ exists (in $\mathbb{R}^{n}$), how would one show that for example $g*f$ or $f(-x)*g(-x)$ exists?
Would it be enough to say that $f$ and $g$ are both locally integrable? Or that both are continuous and one of them has compact support? If that's the case, does it not automatically follow from the fact that $f*g$ exists?
You can do the first of these through change of variables:
with $y=x-z$, $$(f*g)(x)=\int_{\mathbb{R}^n}f(y)g(x-y)dy=(-1)^n\int_{\mathbb{R}^n}f(x-z)g(z)(-1)^ndz=(g*f)(x).$$
The second one isn't defined because you can't take the convolution of two numbers. Do you mean $f_2*g_2$ where $f_2(x)=f(-x)$ and $g_2(x)=g(-x)$? In that case, with $y=-z$, $$(f_2*g_2)(x)=\int_{\mathbb{R}^n}f_2(y)g_2(x-y)dy=(-1)^n\int_{\mathbb{R}^n}f(z)g(-x-z)(-1)^ndz=(f*g)(-x).$$
The $(-1)^n$ factors come from, first, the substitution and, second, swapping the bounds of the integrals.
You don't need any more assumptions than $f*g$ exists.