Initial Operation:
Given two unit vectors u and v in R^n, a transformation is applied to obtain a new vector u_new, defined as: $$ u_{new} = u + \langle u, v \rangle v $$ where <u,v> represents the dot product of u and v, and we can assume for simplicity that both u and v are unit vectors.
Objective:
I wonder how can I find a vector v_inverse that when applied in a similar transformation to u_new (and then normalized, if needed), returns the original vector u. The transformation is defined by the following equation:
$$ u_{new} + \langle u_{new}, v_{inverse} \rangle v_{inverse} = u $$
Is there any way to prove that v_inverse always\sometimes\never exists?
A suitable $v_{\text{inverse}}$ never exists.
Your transformation can be written in terms of matrix multiplication as follows: $$ u_{\text{new}} = [I + vv^T]u $$ Reversing this transformation corresponds to computing the inverse of the matrix $A = I + vv^T$. I claim that this inverse has the form $B = I - \frac 12 vv^T$. To see that $B = A^{-1}$, it suffices to show that $AB = I$. To that end, $$ \begin{align*} AB &= [I + vv^T][I - \frac 12 vv^T] \\ & = I - I\left(\frac 12 vv^T\right) + vv^TI - \frac 12 vv^Tvv^T \\ & = I - \frac 12 vv^T + vv^T - \frac 12 vv^T = I. \end{align*} $$ This matrix of $A^{-1}$ corresponds to the map $u \mapsto u - \frac 12\langle u,v \rangle v$.
For a suitable $w = v_{\text{inverse}}$ to exist, we must have $A^{-1} = I + ww^T$. However, there can be no vector $w$ such that this holds. Indeed, such a $w$ existing would imply that $$ I + ww^T = I - \frac 12 vv^T \implies\\ ww^T = -\frac 12 vv^T. $$ One way to show that these are necessarily distinct matrices is note that the left side has a non-negative trace and the right has a negative trace. We have $$ \operatorname{Tr}(ww^T) = w^Tw = \|w\|^2 \geq 0, \\ \operatorname{Tr}(-\frac 12 vv^T) = - \frac 12 \operatorname{Tr}(vv^T) = - \frac 12 \|v\|^2 = -\frac 12 < 0. $$
It is perhaps noteworthy that if you allow complex vectors and use a bilinear (rather than sesquilinear) "dot-product" over $\Bbb C^n$, then you have the solution $$ v_{\text{inverse}} = \frac{i}{\sqrt{2}} v, $$ where $i^2 = -1$.