Security of a MAC

Let $(Mac, Vrfy)$ be a safe defined $MAC$ over $(K,M,T)$ where $M = \{0,1\}^n$ and $T = \{0,1\}^{128}$.
Is the following $MAC$ safe? Show your proof.

$Mac'(k,m) = Mac(k, m ⊕ m)$
$Vrfy'(k,m,t) = Vrfy(k,m ⊕ m, t)$

EDIT:
So I tried something, I don't know if it is correct tho, can you please confirm it?
So, because $m⊕m = 0$ the $Mac'(k,m) = Mac(k, m ⊕ m)$ produces a tag $t$ for $0$ and $k$ key, it means that $Vrfy'(k,m,t) = Vrfy(k,m ⊕ m, t)$ will always return $true$ because it verifies the same $m⊕m$ message which is $0$ and $k$ key, which is equal to the returned $t$ ? is that correct?

• Is there an efficient attack on it? ​ If no, why? ​ ​ ​ ​ – user991 Jun 7 '16 at 12:15
• Hint: what is $m \oplus m$? – poncho Jun 7 '16 at 12:17
• @poncho the sent message XOR sent message ? What exactly do you mean by that question? Sorry, I really don't understand what I have to be looking for, I am a newbie and I have an exam in 3 days and I need to understand this type of problems. – southpaw22 Jun 7 '16 at 12:25
• And when you xor anything with itself, what is the result? – poncho Jun 7 '16 at 12:29
• @poncho XOR-ing something with itself gives you $0$, right? so $m⊕m = 0$ ? Is that right ? – southpaw22 Jun 7 '16 at 12:45

For $Mac'$, what the attacker could do is select a message (for example, the string $0$), and query the Oracle for the corresponding tag $T$. So, the attacker knows that $0, T$ is a valid message/tag pair (but that doesn't count for the goal, as he asked for that message).
So, the attacker selects a different message for the same length (for example, the string $1$), and form the message pair $1, T$ (where $T$ is the tag he got previously). As $Vrfy'(k, 1, T) = Vrfy(k, 1 \oplus 1, T) = Vrfy(k, 0 \oplus 0, T) = Vrfy'(k, 0, T)$ evaluates to true (as $0, T$ is a valid pair), so is $1, T$ as well (and so the attacker wins, as he has found the MAC for a message he didn't ask for)