Average square gain of a matrix over all possible vectors

Equip $S^{n - 1} = \{x \in \mathbb{R}^n : |x| = 1\}$ with it's usual surface measure, normalized to be a probability measure. Then you seek $\int_{S^{n - 1}}|Ax|^2\,dS(x)$.

As for selecting parameters, any parameterization of $S^{n - 1}$ will work. You have to apply the formula for the surface integral though:

If $\phi : O \to U \subset {S}^{n - 1}$ is a $C^1$ parameterization and $f : S^{n - 1} \to \mathbb{R}$ vanishes off $U$, then $$\int_{U}f(x)\,dS(x) = \frac{1}{\text{Area}(S^{n - 1})}\int_{O}f(\phi(x))\sqrt{\det D\phi(x)^T D\phi(x)}\,dx.$$ I think you forgot the $\sqrt{ \det D\phi(x)^T D\phi(x)}$. Luckily, this problem is simple enough that we don't need to use coordinates.

Write the SVD of $A$ as $A = UDV^T$. Then $|Ax|^2 = |DV^Tx|^2$. So by invariance of the measure under orthogonal transformations, $$\int_{S^{n - 1}}|Ax|^2\,dS(x) = \int_{S^{n - 1}}|DV^Tx|^2\,dS(x) = \int_{S^{n - 1}}|Dx|^2\,dS(x).$$ Now note $|Dx|^2 = (D^2x, x) = \sigma_1^2|x_1|^2 + \dots + \sigma_n^2|x_n|^2$. Hence $$\int_{S^{n - 1}}|Dx|^2\,dS(x) = \sum_{i = 1}^{n}\sigma_i^2\int_{S^{n - 1}}|x_i|^2\,dS(x).$$ Note that $a_i = \int_{S^{n - 1}}|x_i|^2\,dS(x)$ is independent of $i$ by invariance of the measure under orthogonal transformations. Moreover, $$a_1 + \dots + a_n = \int_{S^{n - 1}}|x|^2\,dS(x) = 1.$$ Hence $a_i = \frac{1}{n}$. In conclusion, $$\int_{S^{n - 1}}|Ax|^2\,dS(x) = \frac{\sigma_1^2 + \dots + \sigma_n^2}{n}.$$

Edit: Here is why $a_i = \int_{S^{n - 1}}|x_i|^2 dS(x)$ is independent of $i$. Fix $i$. Note that \begin{align} \int_{S^{n - 1}}|x_i|^2\,dS(x) &= \int_{S^{n - 1}}|(x, e_i)|^2\,dS(x). \end{align} Now pick an orthogonal transformation $R$ such that $Re_i = e_1$. Since $(x, e_i) = (Rx, Re_i) = (Rx, e_1)$, it follows that $$\int_{S^{n - 1}}|x_i|^2\,dS(x) = \int_{S^{n - 1}}|(Rx, e_1)|^2\,dS(x) = \int_{S^{n - 1}}|(x, e_1)|^2\,dS(x),$$ Where the last equality follows from invariance of the measure under orthogonal transformations. So $a_i = a_1$.

The definition of the "surface measure" I am using here is the one that works for any $C^1$ $m$-dimensional surface $M \subset \mathbb{R}^k$. The measure $\mu$ is a Borel measure on $M$ defined in such a way that if $f : M \to [0, \infty]$ is a measurable function vanishing off the image $U \subset M$ of a coordinate parameterization $\phi : O \to U$, $O$ open in $\mathbb{R}^m$, then $$\int_{M}f\,d\mu = \int_{U}f\,d\mu = \int_{O}f(\phi(x))\sqrt{g(x)}\,dx, \hspace{20pt} g(x) = \det D\phi(x)^T D\phi(x).$$ Here I used it for $M = S^{n - 1}$ (I also normalized it to $1$). From this you can prove the assertion that the measure on $S^{n - 1}$ is invariant under orthogonal transformations, that is, if $R^TR = I$, and $f : S^{n - 1} \to [0, \infty]$ is measurable, then $$\int_{S^{n - 1}}f(Rx)\,d\mu(x) = \int_{S^{n - 1}}f(x)\,d\mu(x).$$ A way to prove this is to first prove it for $f$ vanishing off of a coordinate patch, in which case it is just a computation. Then the fact that $M$ is always a countable union of such patches implies the general case.

The above measure is also defined on a general Riemannian manifold $M$, in which case $D\phi(x)^T D\phi(x)$ is replaced with the coordinate matrix $G(x)$ of the metric tensor of $M$ (the above definition is the special case when the metric tensor is the dot product).