1. Use the condition $g(f(x))=x$

Since $f(x)=\alpha x^5+\beta x^3+\gamma x,$ with $\alpha,\beta,\gamma>0$, we have $f'(x)=5\alpha x^4+3\beta x^2+\gamma>0 \quad \forall x\in\mathbb R.$ So $f$ is strictly increasing on $\mathbb R$, hence one-one and onto (being an odd polynomial of odd degree). Therefore $g$ is the inverse function of $f$, so $f(g(y))=y$ for every real $y$ in the range of $f$, and the range of $f$ is all of $\mathbb R$.

Thus, $f\left(g\left(\frac1n\sum_{i=1}^n f(a_i)\right)\right)=\frac1n\sum_{i=1}^n f(a_i).$ So the problem reduces to finding $\frac1n\sum_{i=1}^n f(a_i).$

2. Use the fact that the terms are in A.P. with mean zero

Given $a_1,a_2,\dots,a_n$ are in arithmetic progression and their mean is zero, $\frac1n\sum_{i=1}^n a_i=0.$

Let the A.P. be written as $a_i=a+(i-1)d.$ Since its mean is zero, the sequence is symmetric about $0$.

Now note that $f(x)=\alpha x^5+\beta x^3+\gamma x$ is an odd function, because it contains only odd powers of $x$. Hence $f(-x)=-f(x).$

For an A.P. with mean zero, terms occur symmetrically as $x,-x$ (and possibly one middle term $0$ if $n$ is odd). Therefore, $\sum_{i=1}^n a_i=0, \quad \sum_{i=1}^n a_i^3=0, \quad \sum_{i=1}^n a_i^5=0.$ So,

Hence, $\frac1n\sum_{i=1}^n f(a_i)=0.$

3. Substitute back

Therefore, $f\left(g\left(\frac1n\sum_{i=1}^n f(a_i)\right)\right)=f(g(0))=0.$ (Equivalently, since $f\circ g$ is identity on $\mathbb R$, this is just the inside quantity.)

4. Check options

The value is $0.$ So the correct option is:

A: 0