Claim: all irreducible representations of a given group, $G$, are pairwise isomorphic.
Proof: Let $[r(g), V]$ and and $[p(g), W]$ be irreducible representations of group $G$. Select a fixed vector $v_0$ in $V$, which is NOT an eigenvalue of $r(g)$ for any $g$, except $**1**$. Define the transformation $T$, that maps vector $r(g)v_0$ (in $V$) to vector $p(g)w_0$ in $W$. Extend this mapping to include any $g$ in $G$.
The map $T$ intertwines the two representations. Consider any group element $h$
$T r(h)r(g)v_0 = T r(hg)v_0 = p(hg)v_0 = p(h)p(g)v_0 = p(h) T r(g)v_0$
Because the representations are irreducible, the map is defined for all of $V$ (i.e. $r(g)v_0$ includes all of $V$, for different $g$ in $G$), and is surjective onto all of $W$. By Schur's lemma, a surjective map between irreducible representations must be an isomorphism.
The best thing to do when confronted with something that seems absurd is to look at an example and see what happens.
Consider $G=S_3$, consider $V$ = sign representation, and $W$ the trivial representation (so they are both one-dimensional and irreducible).
You can immediately see that your first requirement fails here: any vector is an eigenvector for $r(g)$ (which is either the identity or multiplication by $-1$), so you cannot select your $v$ (this means that you'd have to take out the "all irreducible representations" above, maybe you meant "all irreducible representation of dimension more than one"?).
Moreover, I assume that you meant "is not an eigenvector for any $g \neq 1$", since any nonzero vector is an eigenvector for the identity matrix $r(1)$.
Now consider $G=S_4$. Pick $V$ as the degree two irreducible representation, the one in which the normal $V_4$ acts trivially, and the quotient (isomorphic to $S_3$) acts as the unique $2$-dimensional irreducible representation of $S_3$. Here we have a kernel! So there are other elements that act as the identity, therefore, again, there is no such $v_0$.
We could now go search for "all irreducible faithful representations of dimension more than one"... and we would get a problem when we notice that $T$ cannot be extended to all $g \in G$. (what is $w_0$, by the way? Same as $v_0$ but for $W$?) but before elaborating on that, are you sure this is what you wanted?