For $u\in H:=H^1(\mathbb R^N)$ we consider $I(u)=\frac12|\nabla u|_2^2-\mu |u|_p^p$ for some fixed $\mu>0$ and $p\in (2,2^*)$ where $2^*$ is the Sobolev critical value and $|~\cdot~|_p$ is the norm from $L^p(\mathbb R^N)$.
Further, there is a free action $*:(0,\infty)\times H\to H$ with $1*u=u$ for all $u\in H$.
Since we are interested in critical points of $I$, we introduce the Pohozaev set $$ \mathcal P=\{u\in H~:~\partial_t I(t*u)\mid_{t=1}=0\}. $$
Now, I like to find a continuous path from $\hat u\in\mathcal P$ to $\tilde u\in\mathcal P$ wrt the $H$-norm.
Obviously there is a continuous path $\gamma:[0,1]\to H$ such that $\gamma(0)=\hat u$ and $\gamma(1)=\tilde u$. And further, I proved that for each $u\in H$ there exists a unique $t_u\in(0,\infty)$ such that $t_u*u\in \mathcal P$. Hence, I can define $$ \pi:H\to \mathcal P,~u\mapsto t_u*u. $$
If $\pi$ is continuous wrt the norm in $H$, then $\pi\circ \gamma$ is a continuous path within $\mathcal P$.
I'm not totally sure it $u\mapsto t_u$ is continuous, because $t_u$ depends on $|u|_p$. For $u,v\in H$ close to each other wrt $H$-norm the difference $||u|_p-|v|_p|$ might be large such that $|t_u-t_v|$ might be large too. Therefore, I think that $\pi$ don't have to be continuous and my argument fails.
I found a proof to my claim.
First, we prove: If $v$ is infinitely smooth with compact support, then the map $t\mapsto t*v$ is continuous wrt $H$-norm.
Next, we conclude, using the density of infinitely smooth functions with compact support in $H$:
If $u\in H$, then the map $t\mapsto t*v$ is continuous wrt $H$-norm.
Obviously $t_u$ depends continuously on $u$ because $$ \left||u|_p-|v|_p\right|\leq |u-v|_p\leq c\|u-v\|_{H} $$ for some $c>0$ because $H$ is continuously embedded in $L^p(\mathbb R^N)$.
Finally, the composition of the continuous functions $u\mapsto t_u$ and $t\mapsto t*u$ gives the continuous function $u\mapsto t_u\mapsto t_u*u$.