Let these numbers be $a,b,c \in \mathbb{R}^+$, then $a+b^2+c^2 = a^2+b+c^2=a^2+b^2+c$. Considering these equalities two at a time, we get the following three equations: $a+b^2=a^2+b \iff 0 = (a-b)(a+b-1)$ $b + c^2 = b^2+c \iff 0=(b-c)(b+c-1)$ $a^2+c=c^2+a \iff 0=(c-a)(c+a-1)$ If $a-b,b-c,c-a \neq 0$, then $a+b-1=b+c-1=c+a-1=0 \implies a=b=c=\frac{1}{2}$. If two or more of $a-b,b-c,c-a$ are $0$, then it trivially follows that $a=b=c$. However, if only one of $a-b,b-c,c-a$ is $0$, then we get infinitely many triples where all three numbers are not the same, such as those of the form $(a,b,c)=(t,t,1-t)$ with $0<t<\frac{1}{2}$ or $\frac{1}{2}<t<1$. Therefore, $\boxed{\text{no}}$, not all numbers are always the same.