To find the coefficient of \(x\) in the expansion of \(\left( \frac{1}{ax} + 2ax^2 \right)^5\), we use the binomial theorem. The general term in the expansion is given by:

\(T_k = \binom{5}{k} \left( \frac{1}{ax} \right)^{5-k} (2ax^2)^k\)

We need the term where the power of \(x\) is 1. The power of \(x\) in the general term is:

\(-5 + k + 2k = -5 + 3k\)

Setting \(-5 + 3k = 1\), we solve for \(k\):

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) into the general term:

\(T_2 = \binom{5}{2} \left( \frac{1}{ax} \right)^{3} (2ax^2)^2\)

\(= 10 \times \frac{1}{a^3x^3} \times 4a^2x^4\)

\(= 10 \times \frac{4a^2x^4}{a^3x^3}\)

\(= 10 \times \frac{4a^2x}{a^3}\)

\(= \frac{40a^2x}{a^3}\)

\(= \frac{40x}{a}\)

The coefficient of \(x\) is \(\frac{40}{a}\). We are given that this coefficient is 5:

\(\frac{40}{a} = 5\)

Solving for \(a\):

\(40 = 5a\)

\(a = 8\)