I read a theorem
If $A$ is an invertible square matrix, then $\operatorname{adj} A^T= (\operatorname{adj} A)^T$.
But after attempting to prove it myself and also reading the proof I am unable to tell why the function needs to be invertible for above to be true. As in the proof we no where use this condition.
Proof:
$$\begin{align} A \operatorname{adj}A & =|A|I_n \\ \implies (A \operatorname{adj} A)^T & = \{|A|I_n\}^T\\ \implies (\operatorname{adj} A)^T(A^T) & =|A|I_n \tag{1}\\ \text{Also,}\\ (\operatorname{adj} A^T)(A^T) & = |A^T|I_n \\ (\operatorname{adj} A^T)(A^T) & = |A|I_n \tag{2}\\ \text{from (1) and (2), we get} \\ (\operatorname{adj}A^T)(A^T)& =(\operatorname{adj} A)^T(A^T)\\ \implies \operatorname{adj} A^T&=(\operatorname{adj} A)^T \end{align}$$
You have already used $A$ invertible in the last step, where you removed $A^T$ on the right on both sides.
But you are right that this theorem does not need A to be invertible. By definition $$(\mathbb{adj}A)^T = \begin{bmatrix}A_{11} & A_{21} & \cdots & A_{n1}\\A_{12}&A_{22}&\cdots&A_{n2}\\\vdots&\vdots&\ddots&\vdots\\A_{1n}&A_{2n}&\cdots&A_{nn}\end{bmatrix}^T=\begin{bmatrix}A_{11} & A_{12} & \cdots & A_{1n}\\A_{21}&A_{22}&\cdots&A_{2n}\\\vdots&\vdots&\ddots&\vdots\\A_{n1}&A_{n2}&\cdots&A_{nn}\end{bmatrix}=\begin{bmatrix}{A^T}_{11} & {A^T}_{21} & \cdots & {A^T}_{n1}\\{A^T}_{12}&{A^T}_{22}&\cdots&{A^T}_{n2}\\\vdots&\vdots&\ddots&\vdots\\{A^T}_{1n}&{A^T}_{2n}&\cdots&{A^T}_{nn}\end{bmatrix}=\mathbb{adj}(A^T)$$
Here $A_{ij} = (-1)^{i+j}\mathbb{det}B_{ij},$ where $B_{ij}$ is the $(n-1)\times(n-1)$ matrix you get when you remove the i-th row and the j-th column of $A$. And ${A^T}_{ji} = (-1)^{i+j}\mathbb{det}C_{ji},$ where $C_{ji}$ is what you get when you remove the j-th row and the i-th column of $A^T$. You could directly check that $B_{ij}=(C_{ji})^T$, which implies thay have the same determinant, and hence $A_{ij}={A^T}_{ji}$.