$a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.

We have $$a^4\le a^4+b^4+c^4+d^4=16\implies a\le 2$$ So, we can have $a^4(a-2)\le 0$, i.e. $$a^5\le 2a^4$$ Similarly, $$b^5\le 2b^4,\quad c^5\le 2c^4,\quad d^5\le 2d^4$$ giving $$a^5+b^5+c^5+d^5\le 2(a^4+b^4+c^4+d^4)=32$$ Note here that $$a^5+b^5+c^5+d^5=2(a^4+b^4+c^4+d^4)$$ holds when $$a^4(a-2)=b^4(b-2)=c^4(c-2)=d^4(d-2)=0$$

Suppose $a,b,c,d\geq 0$ and set $x=a^4,\,y=b^4,\,z=c^4,\,w=d^4$. Then $$f(x,y,z,w)=x^{5/4}+y^{5/4}+z^{5/4}+w^{5/4}$$ is strictly convex and thus restricted to the domain $$\{x,y,z,w\geq 0: x+y+z+w=16\}$$ it attains the maximum only in the extremal points, which are the four vertices $(16,0,0,0)$ and cyclicals.

If one number is less then zero than taking the moduli the value of $a^5+b^5+c^5+d^5$ strictly increases.

Note that (by dividing the equations by $16$ and $32$) this is equivalent to showing one of $a,b,c,d$ is $1$ and the others are $0$ if $a^4 + b^4 + c^4 + d^4 = 1$ and $a^5 + b^5 + c^5 + d^5 = 1$. If none of $a,b,c,d$ are $1$ then they must all have absolute value less than one (otherwise the sum of their fourth powers would exceed $1$). It then suffices to note that

$a^5 + b^5 + c^5 + d^5 < a^4 + b^4 + c^4 +d^4 = 1,$

since $a^5 < a^4, b^5 < b^4, c^5<c^4, d^5<d^4$.

Therefore $a^5 + b^5 + c^5 + d^5 \neq 1$ and we can deduce the result.