Is there another N×N matrix that works like **I**?

No. This is not quite obvious, but since 1 is unique you might
suspect that **I** is unique.

It is easy to show that this is so.
Suppose you had a matrix **Z** and that

(1)ZA=A

for any **A**.
We know how **I** works:

(2)BI=B

for any **B**.
Now use (1) with **I** replacing **A**:

(3)ZI=I

Now use (2),
substituting **Z** for **B**:

(4)ZI=Z

Looking at (3) and (4):

(4)Z=I

It is also true that it works with the other
order: **AZ** = **A**.