Given that, ${\left(x^2+5x+5\right)}^{x+5}=1$

This is possible when $x^2+5x+5\ne 0 \& x+5=0$ or $x^2+5x+5=1$ or $x^2+5x+5=-1 \& x+5=\mathrm{even} \mathrm{integer}$

Case1, $x+5=0$

$\Rightarrow x=-5$

And for this value of $x$, $x^2+5x+5\ne 0$

Case2, $x^2+5x+5=1$

$\Rightarrow x^2+5x+4=0$

$\Rightarrow \left(x+1\right)\left(x+4\right)=0$

$\Rightarrow x=-1, -4$

Case3, $x^2+5x+5=-1$

$\Rightarrow x^2+5x+6=0$

$\Rightarrow \left(x+2\right)\left(x+3\right)=0$

$\Rightarrow x=-2, -3$

But for $x=-2,$ we get $x+5=3$ which is not even.

Hence, the number of integers satisfying this equation is $4$.