Can blocks in AES-CBC ciphertext be manipulated in such a way that plaintext becomes shuffled?
It does not seem like this is possible. Consider the simple case where a 2 block (32 byte) plaintext message has been encrypted using AES-CBC with key $k$ and initialization vector $IV$. Let the plaintext blocks be referenced as $P_1$ and $P_2$, and let the corresponding ciphertext blocks be referenced as $C_1$ and $C_2$. In this case, knowing that $\text{AES.encrypt}(k, P_1 || P_2) = C_1 || C_2$ and that $\text{AES.decrypt}(k, C_1 || C_2) = P_1 || P_2$, you are trying to find out whether $\text{AES.decrypt}(k, f(C_1) || f(C_2))$ can possibly equal $P_2 || P_1$, where $f$ is some manipulating function applied to the ciphertext.

From the regular CBC decryption of $(C_1 || C_2)$ into $(P_1 || P_2)$, we are guaranteed the following relationships:

$\text{AES.decrypt}(k, C_2) \oplus C_1 = P_2$

$\text{AES.decrypt}(k, C_1) \oplus IV = P_1$

In order for $(f(C_1) || f(C_2))$ to decrypt into $(P_2 || P_1)$, the following relationships must be true:

$\text{AES.decrypt}(k, f(C_2)) \oplus f(C_1) = P1$

$\text{AES.decrypt}(k, f(C_1)) \oplus IV = P_2$

And then, by substitution, the following must also be true:

$\text{AES.decrypt}(k, f(C_2)) \oplus f(C_1) = \text{AES.decrypt}(k, C_1) \oplus IV$

$\text{AES.decrypt}(k, f(C_1)) \oplus IV = \text{AES.decrypt}(k, C_2) \oplus C_1$

In order for the first equation to be true, $f(C_2)$ must equal $C_1$ and $f(C_1)$ must equal $IV$, and in order for the second to be true as well, $f(C_1)$ must equal $C_2$ and $IV$ must equal $C_1$, giving us the following:

$f(C_2) = C_1$

$f(C_1) = C_2$

$f(C_1) = IV$

$C_1 = IV$

$f(C_1) = C_1 \rightarrow C_1 = C_2$

So the only way that you can manipulate blocks in the 32-byte AES-CBC output to decrypt to rearranged blocks of the original plaintext is if the output blocks $C_1$ and $C_2$ are identical. Clearly if this was the case, the corresponding input blocks would also be identical, and there would be no point for a malicious actor to rearrange them in the first place.