Let $\mathbb{S}^{n-1}$ be the unit sphere in $\mathbb{R}^{n}$ and let $f$ be smooth on $\mathbb{R}^{n}$. Is it possible to express the integral
$$I(\lambda):=\int_{\mathbb{S}^{n-1}}f(\lambda x_1,x_2,\cdots,x_n) d\sigma(x),\qquad \lambda >0$$
in terms of the integral $$J:=\int_{\mathbb{S}^{n-1}}f(x_1,x_2,\cdots,x_n) d\sigma(x).$$
This is a very simple special case of the my old question here Changing variables in integration over spheres
Some thought: If the point $(x_1,x_2,\cdots,x_n)$ lives on the sphere $x_1^2+\cdots+x_n^2=1$ then $(\lambda x_1,x_2,\cdots,x_n)$ lives on the ellipsoid $E_{\lambda}:=\{(x_1,x_2,\cdots,x_n)\in \mathbb{R}^n: x_1^2/\lambda^2+\cdots+x_n^2=1\}$.
So, I am guessing
$$I(\lambda):=\int_{E_{\lambda}}f( x_1,x_2,\cdots,x_n) d\tilde{\sigma}(x)\qquad (1)$$
where $d\tilde{\sigma}$ is the surface measure on the ellipsoid $E_\lambda$.
An edit:
Thanks to the comments below I realize the question so put does not make sense. So, I modify: Can we express $I(\lambda)$ in terms of $J$ where $$J:=\int_{\mathbb{S}^{n-1}}f(x_1,x_2,\cdots,x_n) d\widetilde{\sigma}(x)$$ where $d\widetilde{\sigma}$ is some kind of a weighted surface measure on the sphere, obviously realted to the natural measure $d{\sigma}$ ?
The answer is no, even with the modification: The "stretched" or "squashed" spheres (gray) are mutually disjoint except for a set of measure zero (the equatorial $S^{n-2}$ where $x_{1} = 0$, shown in blue). Since there's no constraint on $f$, there's no control over the integrals $I(\lambda)$.
The only hope we have in this type of situation, possibly using a suitable $\lambda$-dependent measure on the unit sphere, is when working with functions $f$ whose value on the sphere determines the values on the transformed set.
(One other tangential point: If we integrate the constant function $1$ over the unit circle, we get the arc length $2\pi$. If we integrate the same constant function over a stretched sphere, we may well obtain the arc length of an ellipse, a non-elementary function of $\lambda$.)