For example, we had such an equation (where A is a matrix, and we need to find X, also where T is transpose) -
A = $(2 \cdot X)^T$
So, we will have something like this? According to the sources which I found.
$$A = X^T \cdot 2 \cdot I^T$$
$$X^T = A \cdot (2 \cdot I^T)^{-1}$$
Scalars, such as 2 are not effected by transposition, since transposing only makes rows columns and the other way around, so multiplying all entries of the matrix with two before or after transposing doesn't make a difference: $$ A = (2 X)^T = 2 X^T. $$ More precisely one could write $$ (2 X)^T = (2 I \cdot X)^T = X^T \cdot (2 I)^T = X^T \cdot 2 (I)^T = X^T \cdot 2 (I) = 2 X^T, $$ but we still need the above demonstrated fact $(\lambda A)^T = \lambda A^T$.
Solving for $X$, we have $$ A = 2 X^T \implies X^T = \frac{1}{2} A \implies \left(X^T\right)^T = \left( \frac{1}{2} A \right)^T \implies X = \frac{1}{2} A^T. $$
Also, your way is correct, since your last result simplifies as follows. $$ X^T = A \cdot (2 \cdot I^T)^{-1} = A \cdot 2^{-1} I^{-T} = \frac{1}{2} A, $$ which is my second step from above.