We need to solve $\left(x^2-5x+5\right)^{x^2+4x-60}=1.$

Let $a=x^2-5x+5,\qquad b=x^2+4x-60.$ Then the equation is $a^b=1.$

For real numbers, $a^b=1$ happens in the following standard cases:

1. $a=1$
2. $b=0$ and $a\neq 0$
3. $a=-1$ and $b$ is an even integer

We now check each case.

1. Case: $a=1$

So, $x^2-5x+5=1$ $x^2-5x+4=0$ $(x-1)(x-4)=0$ Hence, $x=1,4.$

Both are valid.

2. Case: $b=0$

So, $x^2+4x-60=0$ $(x+10)(x-6)=0$ Hence, $x=-10,6.$

We must ensure $a\neq 0$.

• For $x=-10$: $a=100+50+5=155\neq 0$
• For $x=6$: $a=36-30+5=11\neq 0$

So both are valid.

3. Case: $a=-1$ and $b$ even integer

First solve $x^2-5x+5=-1$ $x^2-5x+6=0$ $(x-2)(x-3)=0$ Hence, $x=2,3.$

Now check whether $b$ is an even integer.

• For $x=2$: $b=4+8-60=-48$ which is an even integer. So $(-1)^{-48}=1$ is valid.

• For $x=3$: $b=9+12-60=-39$ which is odd, so $(-1)^{-39}=-1\neq 1$. Not valid.

Thus only $x=2$ works from this case.

Collect all real solutions

The real values of $x$ are $x=1,4,-10,6,2.$

Their sum is $1+4-10+6+2=3.$

Final Answer

$\boxed{3}$ So the correct option is C.