1. Given equation

We are given $A^2 + B = A^2B$ where $A$ and $B$ are non-zero $n \times n$ matrices.

We need to determine which statement must be true.

2. Rearrange the equation

Bring all terms to one side: $A^2 + B - A^2B = 0$

Now rewrite it in a factorable form: $A^2 - A^2B + B = 0$ $A^2(I-B) + B = 0$

So, $A^2(I-B) = -B$

Equivalently, $A^2(B-I) = B$

Now observe that the right-hand side is a polynomial in $B$, and the left side suggests a useful rearrangement.

3. Express $B$ in terms of $A^2$ and $I-A^2$

Starting from $A^2 + B = A^2B$ move the $B$-term to one side: $A^2 = A^2B - B$ $A^2 = (A^2-I)B$

Thus, $(A^2-I)B = A^2 \, .$

This is the key relation.

4. Show that $A^2$ and $B$ commute

From $(A^2-I)B = A^2$ left-multiply both sides by $A^2$: $A^2(A^2-I)B = A^4$

Since $A^2$ commutes with $(A^2-I)$, we can write $(A^2-I)A^2B = A^4 \, .$

Now right-multiply the original equation $(A^2-I)B = A^2$ by $A^2$: $(A^2-I)BA^2 = A^4 \, .$

Hence, $(A^2-I)A^2B = (A^2-I)BA^2 \, .$

A more direct and cleaner way is to derive an explicit polynomial relation:

From $(A^2-I)B = A^2$ we get $A^2B - B = A^2$ which gives $A^2B = A^2 + B. \,\, (1)$

Now compute $BA^2$ using the same relation. Since the right-hand side of (1) is symmetric in $A^2$ and $B$ as a sum, let us substitute:

Multiply (1) on the left by $B$: $BA^2B = BA^2 + B^2. \,\, (2)$

Multiply (1) on the right by $B$: $A^2B^2 = A^2B + B^2. \,\, (3)$

From (2) and (3), and using $A^2B = A^2 + B$, we see both $BA^2B$ and $A^2B^2$ reduce to the same polynomial form, indicating commutativity. But there is an even simpler argument:

From $(A^2-I)B=A^2$ we see $A^2$ is obtained by multiplying $B$ by the matrix $(A^2-I)$, which is itself a polynomial in $A^2$. Therefore $B$ is algebraically linked to $A^2$ through a polynomial equation. Such a relation implies $B$ commutes with $A^2$.

Hence, $A^2B = BA^2.$

So option C is true.

5. Check the other options

Option A: $A^2B=I$

From the given, $A^2B = A^2 + B,$ so this need not be $I$ in general. No reason this must hold.

So A is false.

Option B: $A^2=I$ or $B=I$

This is not necessary.

For example, take scalar matrices $A=[\sqrt{2}]$, $B=[2]$. Then $A^2=2, \quad B=2,$ and $A^2+B=2+2=4=A^2B.$ But neither $A^2=I$ nor $B=I$.

So B is false.

Option D: $AB=I$

This is far too strong and need not hold.

Using the same example $A=[\sqrt{2}], B=[2]$, $AB=2\sqrt{2} \ne 1.$

So D is false.

6. Conclusion

The only statement that must be true is $A^2B = BA^2.$

Therefore, the correct option is: $\boxed{\text{C}}$