Of course, as previously noted by other users, you cannot transpose
a matrix by a simple multiplication by a scalar. Transposition is an
operation of its own.
The complesity of matrix operations depends on the way the matrices are represented.
The cost of matrix operations is very dependent on the way matrices
are actually represented in your algorithm. While the most obvious
representations can have a cost $O(nm)$ (or $O(n^2)$ for a square matrix) for scalar multiplication or
for transposition, other representations can be chosen, depending on
the algorithm it is used for, with a lower cost.
The complexity of multiplying a matrix by a scalar $\alpha$ in the
usual way does imply multiplying each of its $n\times m$ elements by
$\alpha$, and hence has a cost $O(nm)$. However, in general, the cost
depends on the way you actually represent matrices, and it is
conceivable to use representations where the cost might be constant,
if the matrix is represented up to scalar multiplication together with
a scalar factor. The computational cost can also be lower in the case of a sparse matrix representation that allows iterating only on the useful (or non-zero) elements.
Without going into details, matrix transposition can
also be done in constant time, i.e., complexity $O(1)$, if you choose to represent a matrix as
a pair composed of an appropriate memory structure of elements, and an indexation function for accessing them,
given the row and column number. Then transposition can be achieved in
constant time simply by changing the indexation function.
Note: this is a matter of representation, not of model of
computation (the expression computational model being ambiguous).
But I am not sure what is intended by Tom van der Zanden'comment.