The correct answer is $\boxed{\frac{4}{5}}$.
To solve this, we can use the formula for the inverse of a 2×2 matrix:
$$\left[ {\begin{array}{{20}{c}} {a}&{b} \ {c}&{d} \end{array}} \right]^{-1} = \frac{1}{ad-bc} \left[ {\begin{array}{{20}{c}} {d}&{-b} \ {-c}&{a} \end{array}} \right]$$
In this case, we have $a=\frac{3}{5}$, $b=\frac{4}{5}$, $c=x$, and $d=\frac{3}{5}$. Substituting these values into the formula, we get:
$$\left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{\frac{4}{5}} \ {x}&{\frac{3}{5}} \end{array}} \right]^{-1} = \frac{1}{\frac{3}{5}\times\frac{3}{5}-\frac{4}{5}x} \left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{-\frac{4}{5}} \ {-x}&{\frac{3}{5}} \end{array}} \right]$$
Simplifying, we get:
$$\left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{\frac{4}{5}} \ {x}&{\frac{3}{5}} \end{array}} \right]^{-1} = \frac{5}{9-4x} \left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{-\frac{4}{5}} \ {-x}&{\frac{3}{5}} \end{array}} \right]$$
We are given that [M]T = [M]-1. This means that the transpose of the matrix is equal to the inverse of the matrix. We can write this as:
$$\left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{\frac{4}{5}} \ {x}&{\frac{3}{5}} \end{array}} \right]^T = \frac{5}{9-4x} \left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{-\frac{4}{5}} \ {-x}&{\frac{3}{5}} \end{array}} \right]$$
Multiplying both sides by the 2×2 identity matrix, we get:
$$\left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{\frac{4}{5}} \ {0}&{\frac{3}{5}} \end{array}} \right] = \frac{5}{9-4x} \left[ {\begin{array}{{20}{c}} {\frac{3}{5}}&{-\frac{4}{5}} \ {-x}&{\frac{3}{5}} \end{array}} \right]$$
Equating the corresponding entries, we get:
$$\frac{3}{5} = \frac{5}{9-4x} \cdot \frac{3}{5}$$
$$\frac{3}{5} = \frac{3}{9-4x}$$
$$9-4x = 5$$
$$4x = 4$$
$$x = \frac{4}{4} = \boxed{\frac{4}{5}}$$