From the definition of the Fréchet derivative and the linearity of the transpose it’s clear that the derivative of the vector transpose is the vector transpose itself.
$$f(x + h) = f(x) + D(x)h + o(\lvert\lvert h \rvert\rvert)$$
$$(x + h)^T = x^T + h^T + 0 = x^T + h^T + o(\lvert\lvert h \rvert\rvert)$$
According to the answer to question, the vector transpose is a linear transformation whose matrix representation is the identity matrix $I$, which is a constant so $D(x) = I$
But $$(x+h)^T \ne x^T + Ih + o(\lvert\lvert h \rvert\rvert)$$
What am I missing?
I think I’ve realized what’s going on. The notation makes sense when the vector $h$ and the matrix $I$ are written out in terms of basis vectors.
$$ h = \alpha_1e_1 + \dots \alpha_ne_n$$
Since the transpose takes $e_i$ to $e_i^T$ and since the entries of the matrix representation of a linear transformation are the inner products between the target basis vectors and the vectors that result from the action of the transformation on the source basis vectors. $$ I = \begin{bmatrix} e_1^T \cdotp (e_1)^T & \dots & e_1^T \cdotp (e_n)^T \\ \vdots & \ddots & \vdots \\ e_n^T \cdotp (e_1)^T & \dots & e_n^T \cdotp (e_n)^T \end{bmatrix}$$
So, for the transpose $$ Ih = I(\alpha_1e_1 + \dots \alpha_ne_n) = \alpha_1e_1^T + \dots \alpha_ne_n^T = h^T$$
It’s probably less confusing to label $I$ as $I_{^T}$ to stress that the action of this matrix transposes the basis vectors.