I think the solution is as follows
$\# \mapsto \text^$
$@ \mapsto +$
$\\\$ \mapsto \times$
$ \% \mapsto -$
$a = -1$
$b = 0$
$c = -2$
$d = -3$
$e=3$.
$f$ can be anything.
$|g| < 6$
$h=1$
$i$ is an odd integer.
$\bigstar =10$
What follows is the explanation of how I found this. Apologies for the length, it is essentially a lot of case-bashing. The solution is found at the very end. Apologies also for any mistakes, there are probably a few in here.
Case 1
$\# \mapsto -$
This would mean that the right hand side of the first inequality is $-a$.
If $@ \mapsto \times$ then the left hand side is non-negative (a square) and so $a$ must be strictly negative. But since $a^2 < -a$, it can only be that $-1 < a< 0$ and this is not allowed.
If $@ \mapsto +$ then the left hand side is $2a$ and it must be that $a$ is strictly negative, we call this case 1.(i).
If $@ \mapsto \text^$ then the left hand side is $a^a$ and again $a$ must be negative, we call this case 1.(ii).
Case 1.(i)
If $ \\\$ \mapsto \times$ then, by equation II, $b \times c = c \times b + b$ and $b=0$.
This would mean $\% \mapsto \text^$ and equation V gives $c^f = a^f - b^f$ but since $b=0$ and $f \neq 0$ it forces $c=-a$ and $f$ must be even (thanks OP).
Equation VIII becomes $a-i = ai^2 - a$ or $a = \frac{i}{2-i^2}$ The right hand side here is an integer only when $i=-2,-1,0,1$ and $2$ and is different to $a$ only when $i=-2$ (where $a=1$) or $i=2$ (where $a=-1$) and since we've established that $a$ is negative, it must be that $a=-1$ and $i=2$ and thus $c=1$.
Looking at equation IV, the right hand side is $e^e - 1$ and the left hand side is either $e$, if $e$ is even or $1/e$, if $e$ is odd. The only integer solution here is $e=0$ but we already have $b=0$ so there is no overall solution in this case.
Alternatively, if $ \\\$ \mapsto \text^$ then equation II gives $b^c = c^b + b$.
A little bit of work can show us that this equation admits just two integer solutions, namely $(b,c) = (0,0)$ (disallowed for being equal) or $(b,c) = (1,0)$. This also means $\% \mapsto \times$ and equation V gives $0 = af - f$ and since $a<0$, it must be that $f=0=c$ which is not allowed so there are no possible solutions in case 1.(i).
Case 1.(ii)
If $ \\\$ \mapsto \times$ then equation II tells us that $bc = cb^b$ which means that either $c=0$ or $b=-1$ or $1$.
This would also mean that $\% \mapsto +$ and equation IV gives $e+a+e = e+e+c^{a^b}$ or $a = c^{a^b}$. This rules out $c=0$ as $a$ would be $0$. If $b=1$ or $-1$ then $c = a^{1/a}$ or $a^a$ but $a=-1$ gives $c=-1$ and other negative values of $a$ make $c$ non integral.
Alternatively, if $ \\\$ \mapsto +$ the equation II becomes $b+c = c+b^b$ which means $b=1$ or $-1$. In this case, $ \% \mapsto \times$ and equation IV tells us that $c = a^{1/a}$ or $a^a$ (which runs into the same problems as previous) or that $e=0$.
If $e=0$, it means that $f \neq 0$ and we can factor it out of equation V to get $c = a-b$. Plugging this into equation III, gives us $d^d = 2a$ but this isn't possible since $a$ is a negative integer and the minimum possible for $d^d$ is $-1$. Hence, there are no possible solutions in case 1.(ii)
Case 2
$\# \mapsto +$
This puts the right hand side of equation I as $3a$.
If $@ \mapsto -$ then the left hand side of I is 0 so we have $a>0$. We call this case 2.(i).
If $@ \mapsto \times$ then we have $a^2 < 3a$ so that $a = 1$ or $2$. We call this case 2.(ii).
If $@ \mapsto \text^$ then we have $a^a < 3a$ so that $a=1$ or $2$. We call this case 2.(iii)
Case 2.(i)
If $ \\\$ \mapsto \times$ then equation II gives $b=0$. But then equation III gives $d^d + a < 0$ which is not possible with $a > 0$.
Alternatively, if $ \\\$ \mapsto \text^$, equation II becomes $b^c = c^b - b$ which has integer solutions $(b,c) = (0,0)$ (forbidden) and $(b,c) = (1,2)$. Looking at equation III, we have $a < c^b$ which is $2$ which puts $a=1=b$. Hence there are no solutions in Case 2.(i).
Case 2.(ii)
If $\\\$ \mapsto -$ then equation VIII gives $3i = a$ which makes $i$ non-integral as $a=1$ or $2$.
Alternatively, if $\\\$ \mapsto \text^$ then equation VIII gives $i = a^{i^i}$ which has no real solutions for $i$ when $a=2$ and gives $i=1$ when $a=1$. Hence, there are no solutions in case 2.(ii)
Case 2.(iii)
Again $\\\$ \mapsto -$ gives us the same issue with $a$ as in 2.(ii).
Alternatively, if $\\\$ \mapsto \times$, equation VIII gives $i=ai^2$ which only has integer solutions $i=1=a$ (disallowed) or $i=0$. However, equation II gives $b=0$ so this does not work. Hence, there are no solutions in case 2.(iii)
Case 3
$\# \mapsto \times$
If $@ \mapsto +$ then it must be that $a>1$, we call this case 3.(i)
If $@ \mapsto -$ then we must have $a>0$, we call this case 3.(ii)
If $@ \mapsto \text^$ then the only possibility is $a=2$, we call this case 3.(iii)
Case 3.(i)
If $\\\$ \mapsto -$ then equation VIII becomes $2ai = a-i$ or $i = \frac{a}{2a+1}$ which is non-integral for $a>1$.
Alternatively, if $\\\$ \mapsto \text^$ then equation VIII becomes $ i = a^{i^i}$ which has no solutions for $a>1$. Hence there are no solutions in case 3.(i).
Case 3.(ii)
If $ \\\$ \mapsto +$ then equation II gives $b=0$. Equation V gives $a^f = c^f$ and since $f \neq 0$ and $a \neq c$ it forces $f$ to be even and $a=-c$. However, equation VIII gives directly $a=-i$, forcing $c=i$ which is not allowed.
Alternatively, if $ \\\$ \mapsto \text^$ then VIII gives $i = a^{i^i}$ which, as before, does not lead to a valid solution. Hence there are no solutions in case 3.(ii)
Case 3.(iii)
If $ \\\$ \mapsto +$ then equation II gives $b = b^b$ meaning that $b=1$ or $-1$.
Equation IV then gives us $a = c^{a^b}$ which makes $c$ irrational if $b=1$, but if $b=-1$ then $c=4$.
However, equation V then tells us that $4-f = (2-f)(-1-f)$ or $f^2 = 6$ which makes $f$ irrational and therefore, no solution in this case.
Alternatively, if $ \\\$ \mapsto -$, equation VIII gives $4i = 2-i$ or $i=2/5$ which is not an integer. Hence, there is no solution in case 3.(iii)
Case 4
$\# \mapsto \text^$
$a<-1$ seems to sometimes create imaginary values on the right hand side of I when $a$ is even but there seem to be valid real numbers in play when $a$ is odd so we'll restrict to these values.
If $@ \mapsto -$ then the left hand side of I is 0 and we just require $a>0$. We call this case 4.(i)
If $@ \mapsto \times$ then the inequality is satisfied only when $a>1$. We call this case 4.(ii)
If $@ \mapsto +$ it appears that $a$ can take any negative odd value and also any value $a>1$. We call this case 4.(iii)
Case 4.(i)
If $\\\$ \mapsto +$ then equation VIII gives $a^i = a + i + i^a$. Any values of $i < -1$ will create a non-integer quantity on the left which cannot be rectified by the integer value of the right hand side. After that, we just need to check through a small number of values for $a$ and $i$ before realising that the largeness of $a^i - i^a$ cannot be compensated by $a+i$. In this case the solutions are $(a,i) = (1,0)$ or $(2,5)$.
Equation II implies $b=0$ so it must be that $a=2$ and $i=5$.
Equation V then gives $cf = 2f$ which means that $c=2=a$ or $f=0=b$ so this case doesn't work.
Alternatively, if $\\\$ \mapsto \times$ the equation II gives $b=0$ and equation IV gives $c=2a$.
Then equation V gives $c=a+1$ so that $a=1$ and $c=2$.
Inequality III then gives $0 < 0$ which is not allowed so there is no solution in case 4.(i)
Case 4.(ii)
If $ \\\$ \mapsto +$ then again equation VIII implies $(a,i) = (2,5)$ (as $(1,0)$ has already been disallowed).
Equation II implies $b=b^2$ so that $b=0$ or $b=1$.
However, equation IV then gives us $bc = 1$ which means $b=c=1$.
Alternatively, if $ \\\$ \mapsto -$ then equation VIII gives $a^i + i^a = a-i$. If $a$ is less than $-1$ then the right hand side of this equation is an integer where the left cannot be, similarly when $i<-1$. If both numbers are greater than $1$, then the left hand side is clearly bigger. The only solution we can get here is thus $a=1, i=0$ but this has already been disallowed. Hence there are no solutions in case 4.(ii)
Case 4.(iii)
If $ \\\$ \mapsto -$ then equation VIII gives $a^i + i^a = a-i$. As before, the only solution we can get here is $a=1, i=0$ but this value for $a$ is disallowed by I.
Alternatively, suppose $\\\$ \mapsto \times$.
Then equation II gives $b=0$. Equation V then gives us $c = a-1$ and equation IV gives $c=2a$. Thus $a=-1$ and $c=-2$.
Equation VIII then gives $(-1)^i = -1$ which tells us $i$ is odd.
Inequality III tells us that $d + d^{-1} < 0$ and so $d$ is negative and necessarily less than $-2$ so as not to cross paths with $a$ and $c$.
In inequality VII, if $h < -1$, then the right hand side is $-\infty$ (something that I missed before) so $h > 0$. Inequality VII then tells us that $c^5 - d^3 < 1$ which means that $d^3 > -33$ and so $d=-3$.
Inequality VI gives $g^2 + d^e < 0$ which means that $e$ is necessarily odd and $>1$ as $e=1$ forces $|g| < 2$ and all values in this range are already taken.
Finally inequality IX gives $e^{2h} < \bigstar < ec^2 + a = 4e-1$.
Since $h > 0$ the only value of $e$ which can make the left smaller than the right is $e=3$ and then we must have $h=1$. Then $e^{2h} = 9$, $4e-1 = 11$ and $\bigstar=10$ is the singular value which works.
