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One can verify that $U_{A}=U_{S}$ as follows. Note that $$\det(U_{A}\Sigma_{A}V_{A}^{T}+U_{S}\Sigma_{S}V_{S}^{T})=\det(\Sigma_{A}+U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T})$$ Let $Y=U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T}$ so that the problem can be rephrased as:

$$\max\limits_{Y} \det(\Sigma_{A}+Y) \text{ subject to } Tr(Y)=c,\ Y\succ 0$$

Now, let $Z=\Sigma_{A}+Y$ so that the problem is reformulated:

$$\max\limits_{Z} \det(Z) \text{ subject to } Tr(Z)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ Z\succ 0,\ Z_{i,i}\geq \sigma_{A}(i)$$

Since $Z$ can be written $Z=UT$ where $U$ is orthogonal and $T$ upper triangular, $\det(Z)=\det(T)$ and the off-diagonal terms of $T$ do not contribute so that $T$ may be assumed diagonal. Thus, finally one arrives at:

$$\max\limits_{T} \det(T) \text{ subject to } Tr(T)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ T\succ 0,\ T_{i,i}\geq \sigma_{A}(i),\ \text{T is diagonal}$$

This problem's solution follows from solving the water filling problem above:

$$\max\limits_{t_{k}} \sum\limits_{k=1}^{n} \log(t_{k}+\sigma_{k}) \text{ subject to } t_{k}\geq 0 \text{ and }\sum\limits_{k=1}^{n} t_{k}=c$$

One can verify that $U_{A}=U_{S}$ as follows. Note that $$\det(U_{A}\Sigma_{A}V_{A}^{T}+U_{S}\Sigma_{S}V_{S}^{T})=\det(\Sigma_{A}+U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T})$$ Let $Y=U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T}$ so that the problem can be rephrased as:

$$\max\limits_{Y} \det(\Sigma_{A}+Y) \text{ subject to } Tr(Y)=c,\ Y\succ 0$$

Now, let $Z=\Sigma_{A}+Y$ so that the problem is reformulated:

$$\max\limits_{Z} \det(Z) \text{ subject to } Tr(Z)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ Z\succ 0,\ Z_{i,i}\geq \sigma_{A}(i)$$

Since $Z$ can be written $Z=UT$ where $U$ is orthogonal and $T$ upper triangular, $\det(Z)=\det(T)$ and the off-diagonal terms of $T$ do not contribute so that $T$ may be assumed diagonal. Thus, finally one arrives at:

$$\max\limits_{T} \det(T) \text{ subject to } Tr(T)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ T\succ 0,\ T_{i,i}\geq \sigma_{A}(i),\ \text{T is diagonal}$$

This problem's solution follows from solving the water filling problem above where $t_{k}=T_{k,k}$:

$$\max\limits_{t_{k}} \sum\limits_{k=1}^{n} \log(t_{k}+\sigma_{k}) \text{ subject to } t_{k}\geq 0 \text{ and }\sum\limits_{k=1}^{n} t_{k}=c$$

Taking $A=\begin{pmatrix} 2 & -1 & 0 \\-1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ and running cvx via MATLAB on the problem replacing the singular value condition with the equivalent trace condition with $c=1$ does roughly give $U_{A}=U_{S}$.

cvx_begin
variable S(3,3) symmetric;
S==semidefinite(3);
minimize(-log_det(S+A));
subject to
trace(S) ==1;
cvx_end

[U,S,V]=svd(A)
[U1,S1,V1]=svd(S)

The output is

Successive approximation method to be employed.
   For improved efficiency, SDPT3 is solving the dual problem.
   SDPT3 will be called several times to refine the solution.
   Original size: 40 variables, 16 equality constraints
   1 exponentials add 8 variables, 5 equality constraints
 -----------------------------------------------------------------
 Cones  |             Errors              |
 Mov/Act | Centering  Exp cone   Poly cone | Status
 --------+---------------------------------+---------
  1/  1 | 2.305e-01  3.496e-03  0.000e+00 | Solved
  1/  1 | 2.716e-02  4.906e-05  0.000e+00 | Solved
  1/  1 | 2.918e-03  5.657e-07  0.000e+00 | Solved
  0/  1 | 3.172e-04  6.511e-09  0.000e+00 | Solved
-----------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -2.38217


ans =

    3.4142
    2.0000
    0.5858


U =

   -0.5000   -0.7071   -0.5000
    0.7071   -0.0000   -0.7071
    0.5000   -0.7071    0.5000


S =

    3.4142         0         0
         0    2.0000         0
         0         0    0.5858


V =

   -0.5000   -0.7071   -0.5000
    0.7071    0.0000   -0.7071
    0.5000   -0.7071    0.5000


U1 =

   -0.5000    0.7068   -0.5004
   -0.7071    0.0004    0.7071
    0.5000    0.7074    0.4996


S1 =

    1.0000         0         0
         0    0.0000         0
         0         0    0.0000


V1 =

   -0.5000    0.7068   -0.5004
   -0.7071    0.0004    0.7071
    0.5000    0.7074    0.4996

Edit: One can verify that $U_{A}=U_{S}$ as follows. Note that $\det(U_{A}\Sigma_{A}V_{A}^{T}+U_{S}\Sigma_{S}V_{S}^{T})=\det(\Sigma_{A}+U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T})$. Let $Y=U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T}$ so that the problem can be rephrased as:

$\max\limits_{Y} \det(\Sigma_{A}+Y) \text{ subject to } Tr(Y)=c,\ Y\succ 0$

$\max\limits_{Z} \det(Z) \text{ subject to } Tr(Z)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ Z\succ 0,\ Z_{i,i}\geq \sigma_{A}(i)$.

$\max\limits_{T} \det(T) \text{ subject to } Tr(T)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ T\succ 0,\ T_{i,i}\geq \sigma_{A}(i),\ \text{T is diagonal}$.

$\max\limits_{t_{k}} \sum\limits_{k=1}^{n} \log(t_{k}+\sigma_{k}) \text{ subject to } t_{k}\geq 0 \text{ and }\sum\limits_{k=1}^{n} t_{k}=c$

One can verify that $U_{A}=U_{S}$ as follows. Note that $$\det(U_{A}\Sigma_{A}V_{A}^{T}+U_{S}\Sigma_{S}V_{S}^{T})=\det(\Sigma_{A}+U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T})$$ Let $Y=U_{A}^{-1}U_{S}\Sigma_{S}V_{S}^{T}V_{A}^{-T}$ so that the problem can be rephrased as:

$$\max\limits_{Y} \det(\Sigma_{A}+Y) \text{ subject to } Tr(Y)=c,\ Y\succ 0$$

$$\max\limits_{Z} \det(Z) \text{ subject to } Tr(Z)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ Z\succ 0,\ Z_{i,i}\geq \sigma_{A}(i)$$

$$\max\limits_{T} \det(T) \text{ subject to } Tr(T)=c+\sum\limits_{i=1}^{n} \sigma_{A}(i),\ T\succ 0,\ T_{i,i}\geq \sigma_{A}(i),\ \text{T is diagonal}$$

$$\max\limits_{t_{k}} \sum\limits_{k=1}^{n} \log(t_{k}+\sigma_{k}) \text{ subject to } t_{k}\geq 0 \text{ and }\sum\limits_{k=1}^{n} t_{k}=c$$

This problem is just the water filling problem above:

This problem's solution follows from solving the water filling problem above: